NCERT Math Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions (English Medium) || CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.3

Here, \( \mathrm{a}=2, \mathrm{~d}=5 \) and \( \mathrm{n}=10 \)
Sum of n terms can be given as follows:
\(S=\frac{N}{2}[2 a+(n-1) d]\)
\(S_{10}=\frac{10}{2}[2 \times 2+(10-1) 5]\)
\(=5(4+45)\)
\(=5 \times 49\)
\(=245\)
thus, sum of the 10 terms of given \( \operatorname{AP}\left(\mathrm{S}_{\mathrm{n}}\right)=245 \)

Here, \( a=-37, d=4 \) and \( n=12 \)
Sum of n terms can be given as follows:
\(S=\frac{N}{2}[2 a+(n-1) d]\)
\(S_{12}=\frac{12}{2}[2 \times(-37)+(11) 4]\)
\(=6(-74+44)\)
\(=6 \times(-30)\)
\(=-180\)
Thus, sum of the 12 terms of given \( A P\left(\mathrm{S}_{\mathrm{n}}\right)=-180 \)

Here, \( \mathrm{a}=0.6, \mathrm{~d}=1.1 \) and \( \mathrm{n}=100 \)
Sum of n terms can be given as follows:
\(S=\frac{N}{2}[2 a+(n-1) d]\)
\(S_{100}=\frac{100}{2}[2 \times(0.6)+(99) .1]\)
\(50(1.2+108.9)\)
\(=50 \times(110.1)\)
\(=5505\)
Thus, sum of the 100 terms of given AP \( (\mathrm{S} 100)=5505 \)

Here,
\(a=\frac{1}{15}, n=11\)
\(d=\frac{1}{12}-\frac{1}{5}=\frac{1}{60}\)
Sum of n terms can be given as follows:
\(S=\frac{N}{2}[2 a+(n-1) d]\)
\(S_{11}=\frac{11}{2}\left[2 \times\left(\frac{1}{15}\right)+(10) \frac{1}{60}\right]\)
\(=\frac{11}{2}\left(\frac{2}{15}+\frac{1}{6}\right)\)
\(=\frac{11}{2} \times\left(\frac{4+5}{30}\right)\)
\(=\frac{33}{20}\)
Thus, sum of the 100 terms of given AP \( \left(\mathrm{S}_{\mathrm{n}}\right)=\frac{33}{20} \)

Here, \( \mathrm{a}=7, \mathrm{~d}=3.5 \) and last term \( =84 \)
Number of terms can be calculated as follows;
\( \mathrm{an}=a+(\mathrm{n}-1) \mathrm{d} \)
Or, \( 84=7+(\mathrm{n}-1) 3.5 \)
Or, \( (\mathrm{n}-1) 3.5=84-7 \)
Or, \( \mathrm{n}-1=\frac{77 }{ 3.5}=22 \)
Or, \( \mathrm{n}=23 \)
Sum of n terms can be given as follows:
\(S=\frac{N}{2}[2 a+(n-1) d]\)
\(S_{23}=\frac{23}{2}[2 \times(7)+(22) .5]\)
\(=\frac{23}{2}(14+77)\)
\(=\frac{23}{2} \times(91)\)
\(=1046 \frac{1}{2}\)

Here, \( \mathrm{a}=34, \mathrm{d}=-2 \) and last term \( =10 \)
Number of terms can be calculated as follows:
\( \mathrm{an}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d} \)
Or, \( 10=34+(n-1)(-2) \)
Or, \( 10=34-(\mathrm{n}-1)(2) \)
Or, \( (\mathrm{n}-1) 2=34-10=24 \)
Or, \( \mathrm{n}-1=12 \)
Or, \( \mathrm{n}=13 \)
Sum of n terms can be given as follows:
\(S=\frac{N}{2}[2 a+(n-1) d]\)
\(S_{13}=\frac{13}{2}[2 \times(34)+(12)(-2)]\)
\(=\frac{13}{2}[68+(-24)]\)
\(=\frac{13}{2} \times(44)\)
\(=286\)
Thus, sum of the given \( \mathrm{AP}(\mathrm{Sn})= 286\)

Here, \( a=-5, d=-3 \) and last term \( =-230 \)
Number of terms can be calculated as follows:
\(\text {an }=a+(n-1) d\)
\(\text {Or, }-230=-5+(n-1)(-3)\)
\(\text {Or, }-230=-5-(n-1) 3\)
Or, \( (\mathrm{n}-1) 3=-5+230=225 \)
Or, \( \mathrm{n}-1=75 \)
Or, \( \mathrm{n}=76 \)
Sum of n terms can be given as follows:
\(S=\frac{N}{2}[2 a+(n-1) d]\)
\(S_{23}=\frac{76}{2}[2 \times(-5)+(-3) 75]\)
\(=\frac{76}{2}(-10-225)\)
\(38 \times(-235)\)
\(=-8930 .\)

To find: \( \mathrm{n}, \mathrm{S}_{\mathrm{n}} \)
Given: \( a_{n}=50 \)
\( a=5 d=3 \)
We know, nth term of an AP is given by
\( a_{n}=a+(n-1) d \)
where \( a \) and \( d \) are first term and common difference respectively and \( n \) are the number of terms of the A.P
Number of terms can be calculated as follows:
\( a_{n}=a+(n-1) d \)
Or, \( 50=5+(\mathrm{n}-1) 3 \)
Or, \( (n-1) 3=50-5=45 \)
Or, \( \mathrm{n}-1=15 \)
Or, \( \mathrm{n}=16 \)
Sum of N terms can be given as follows:
\( S=\frac{N}{2}[2 a+(n-1) d] \)
\( \mathrm{S}_{16}=\frac{16}{2}[2\times(5)+(1 5) 3]\)
\(=8(10+45)\)
\(=8 \times 55\)
\(=440\)
\( \mathrm{S}_{\mathrm{n}}=440 \)

To find: d and \( \mathrm{S}_{13} \)
Given:
\( \mathrm{a}=7 \)
\( a_{13}=35 \)
We know, nth term of an AP is
\( a_{n}=a+(n-1) d \)
where a and d are first term and common difference respectively.
So, Common difference can be calculated as follows:
\( a_{n}=a+(n-1) d \)
Or, \( 35=7+12 d \)
Or, \( 12 \mathrm{d}=35-7=28 \)
Or, \( \mathrm{d}=\frac{7 }{ 3} \)
Sum of n terms can be given as follows:
\(S=\frac{N}{2}[2 a+(n-1) d]\)
\(S_{16}=\frac{13}{2}\left[2 \times(7)+(12) \frac{7}{3}\right]\)
\(=\frac{13}{2}(14+28)\)
\(=\frac{13}{2} \times(42)\)
\(=273\)
\(S_{13}=273\)

(iii) To Find: a, \( \mathrm{S}_{12} \)
Given:
\(\mathrm{d}=3\)
\(\mathrm{a}_{12}=37\)
We know, nth term of an AP is
\( a n=a+(n-1) d \)
where a and d are first term and common difference respectively.
Therefore, First term can be calculated as follows:
\( \mathrm{a}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d} \)
Or, \( 37=\mathrm{a}+11 \times 3 \)
Or, \( \mathrm{a}=37-33=4 \)
\( a=4 \)
Sum of n terms can be given as follows:
\(S=\frac{N}{2}[2 a+(n-1) d]\)
\(S_{12}=\frac{13}{2}[2 \times(4)+(11) 3]\)
\(=6(8+33)\)
\(=6 \times 41\)
\(\mathrm{~S}_{12}=246\)

To find: d, \( a_{10} \)
Given: \( a_{3}=15 \)
\( \mathrm{S}_{10}=125 \)
Sum of n terms can be given as follows:
\(S=\frac{N}{2}[2 a+(n-1) d]\)
\(S_{10}=\frac{10}{2}[2 \times(a)+(9) d]\)
\(125=5(2 \mathrm{a}+9 \mathrm{d})\)
\(25=2 \mathrm{a}+9 \mathrm{d}\quad \ldots\ldots\text{(i)}\)
We know, nth term of an AP is
\( a_{n}=a+(n-1) d \)
where a and d are first term and common difference respectively.
According to the question; the 3rd term is 15, which means;
\( a+2 d=15 \quad \ldots\ldots\text{(ii)}\)
Now,
Subtracting equation 2 times eq(ii) from equation eq(i), we get;
\( (2 a+9 d)-2(a+2 d)=25-30 \)
Or, \( 2 a+9 d-2 a-4 d=-5 \)
Or, \( 5 \mathrm{d}=-5 \mathrm{d}=-1 \)
Now,
Substituting the value of \( d \) in equation (2), we get;
\( a+2(-1)=15 \)
Or, \( a-2=15 \)
Or, \( \mathrm{a}=17 \)
\( 10^{\text {th}} \) term can be calculated as follows;
\(\mathrm{a}_{10}=\mathrm{a}+9 \mathrm{d}\)
\(=17-9=8\)
\(a_{10}=8\)

To find: a and \(a_{9}\)
Given:
\(d=5\)
\(S_{9}=75\)
Sum of n terms can be given as follows:
\(S=\frac{N}{2}[2 a+(n-1) d]\)
\(75=\frac{9}{2}[2 \times(a)+(8) 5]\)
\(75=\frac{9}{2}(2 a+40)\)
\(\frac{50}{3}=2 a \times 40\)
\(a=-\frac{35}{3}\)
Now,
We know, nth term of an AP is
\( a_{n}=a+(n-1) d \)
where a and d are first term and common difference respectively.
\( 9^{\text {th}} \) term can be calculated as follows:
\(a_{9}=a+8 d\)
\(=-\frac{35}{3}+40\)
\( =\frac{85}{3} \)

To find \( : \mathrm{n} \) and an
Given:
\(\mathrm{a}=2\)
\(\mathrm{d}=8\)
\(\mathrm{Sn}=90\)
Sum of n terms can be given as follows
\(S=\frac{N}{2}[2 a+(n-1) d]\)
\(90=\frac{N}{2}[2 \times(2)+(N-1) 8]\)
\(90=\frac{N}{2}(4+8 N-8)\)
\(90=N(2+4 N-4)\)
\(4 N^{2}-2 N-90=0\)
\(2 N^{2}-N-45=0\)
For solving this quadratic equation, we have to factor 1 in such a way that product comes out to be 90 and the difference should be 1 \( (2 \mathrm{N}+9)(\mathrm{N}-5) \)
Hence, \( n=-\frac{9 }{ 2} \) and \( n=5 \)
Rejecting the negative value,
We have \( \mathrm{n}=5\)
We know, nth term of an AP is
\( a_{n}=a+(n-1) d \)
where a and \( d \) are first term and common difference respectively.
Now, \( 5^{\text {th}} \) term will be:
\(a_{5}=a+(5-1) d\)
\(a_{5}=a+4 d\)
\(=2+4 \times 8\)
\(=2+32=34\)
\(\mathrm{a}_{5}=34\)

To find: n and d
Given:
\(a=8\)
\(a_{n}=62\)
\(S_{n}=210\)
Sum of n terms can be given as follows:
\(S=\frac{N}{2}[2 a+(n-1) d]\)
\(210=\frac{N}{2}[a+a+(n-1) 8]\)
Since, \( a_{n}=a+(n-1) d \), we have
\(420=n(8+62)\)
\(420=\mathrm{n} \times 70\)
\(\mathrm{n}=6\)
Now, for calculating d:
we know that \( a_{n}=a+(n-1) d \)
\( a_{6}=a+5 d \)
Or, \( 62=8+5 \mathrm{d} \)
Or, \( 5 \mathrm{d}=62-8=54 \)
Or, \( \mathrm{d}=\frac{54 }{ 5} \)
Common difference, \( d=\frac{54 }{ 5} \)

To find: n and a
Given: \( a_{n}=4 \)
\( d=2 S_{n}=-14 \)
Sum of n terms can be given as follows:
\(S=\frac{N}{2}[2 a+(n-1) d]\)
\(-14=\frac{n}{2}[a+a+(n-1) d]\)
\(-28=n(a+4)\)
\(n=\frac{-28}{a+4} \quad \ldots\ldots\text{(i)}\)
We know;
We know, nth term of an AP is
\( a_{n}=a+(n-1) d \)
where a and d are the first term and common difference respectively.
therefore,
\( 4=a+(n-1) 2 \)
Or, \( 4=\mathrm{a}+2 \mathrm{n}-2 \)
Or, \( a+2 n=6 \)
Or, \( 2 \mathrm{n}=6-\mathrm{a} \)
Or, \( \mathrm{n}=\frac{(6-\mathrm{a}) }{ 2}\quad \ldots\ldots\text{(ii)} \)
Using (i) and (ii):
\( \frac{-28}{a+4}=\frac{6-a}{2} \)
Cross multiplying, we get
\( -56=(6-a)(a+4) \)
\(24+2 a-a^{2}=-56\)
\(a^{2}-2 a-80=0\)
Factorizing 2 in such a way that product of the two terms is 80 and their difference is 2
\( (a+8)(a-10)=0 \)
Therefore, \( \mathrm{a}=-8 \) and \( \mathrm{a}=10 \)
As a is smaller than 10 and d has positive value, hence we'll take \( \mathrm{a}=- 8\)
Now, we can find the number of terms as follows:
\(\mathrm{a}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}\)
\(4=-8+(\mathrm{n}-1) 2\)
\((\mathrm{n}-1) 2=4+8=12\)
\(\mathrm{n}-1=6, \mathrm{n}=7\)
Hence, \( \mathrm{n}=7 \) and \( \mathrm{a}=-8 \)

To find: d
Given: \( \mathrm{a}=3 \)
\( \mathrm{n}=8 \mathrm{S}_{\mathrm{n}}=192 \)
Sum of n terms can be given as follows:
\(S=\frac{N}{2}[2 a+(n-1) d]\)
\(192=\frac{8}{2}[2 \times 3+7 d]\)
\(192=4(6+7 \mathrm{d})\)
\(7 \mathrm{d}=42\)
\(\mathrm{d}=6\)

To Find: a
Given: \( 1=28, \mathrm{S}=144 \)
Sum of n terms can be given as follows:
\(S=\frac{N}{2}[2 a+(n-1) d]\)
\(144=\frac{9}{2}\left[a+a_{n}\right]\)
\(288=\mathrm{n}(\mathrm{a}+28)\)
\(9 \mathrm{a}+252=288\)
\(9 \mathrm{a}=288-252\)
\(9 \mathrm{a}=36\)
\(\mathrm{a}=4\)

Given that, \( \mathrm{a}=9, \mathrm{d}=8 \) and \( \mathrm{S}_{\mathrm{n}}=636 \)
Let no of terms be N.
Now
We know:
\(S=\frac{N}{2}[2 a+(n-1) d]\)
\(636=\frac{N}{2}[2 \times(9)+(N-1) 8]\)
\(\Rightarrow 636=\frac{N}{2}(18+8 N-8)\)
\(\Rightarrow 636=\mathrm{N}(9+4 \mathrm{N}-4)\)
\(\Rightarrow 636=9 \mathrm{N}+4 \mathrm{N}^{2}-4 \mathrm{N}\)
\(\Rightarrow 4 \mathrm{N}^{2}+5 \mathrm{N}-636=0\)
Now to factorize the above quadratic equation, we need to make the product \(( 636.4=2544 ) \) and difference should be \( 5\)
\( \Rightarrow 4 \mathrm{N}^{2}+53 \mathrm{N}- 48 \mathrm{N}-636=0 \) \(\Rightarrow \mathrm{N}(4 \mathrm{N}+53)-12(4 \mathrm{N}+53)=0\)
\(\Rightarrow(\mathrm{N}-12)(4 \mathrm{N}+53)=0\)
Now, \( \mathrm{N}=-\frac{53 }{ 4} \) and \( \mathrm{N}=12 \)
Taking the integral value and rejecting the fractional value, we have \( \mathrm{n}=12 \).

Sum of n terms of an AP is given by,
\( S_{n}=\frac{n}{2}\left(a+a_{n}\right) \)
where 'a' is first term of \( \mathrm{AP}, \mathrm{n} \) is the number of terms of AP and 'an' is last term of AP.
Given, First term, a \( =5 \) and last term, an \( =45 \)
Also, sum \( =400 \) Putting the values in the formula, we get \( 400=\frac{n}{2} \) \(( 5+ 45)\)
\( \Rightarrow 800=50 \mathrm{n} n=\frac{800}{50} \)
\( \Rightarrow \mathrm{n}=16 \).
i.e. there are 16 terms in given AP.
Now the nth term of an \( A P \) is given by the formula,
\( a_{n}=a+(n-1) d \)
where, \( a \) is the first term, \( n \) is the number of terms, \( a_{n} \) is the \( n \)th term and \( d \) is the common difference.
\(45=5+(16-1) \times d \Rightarrow 15 d=40\)
\(\Rightarrow d=\frac{40 }{ 15}\)
Therefore, common difference of AP is \( \frac{8 }{ 3} \).

We have;
First term, \( \mathrm{a}=17 \), Last term, \( \mathrm{an}=350 \) and
Common difference, \( d=9 \)
We know, nth term formula:
\(\mathrm{a}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}\)
\(350=17+(\mathrm{n}-1) 9\)
\((\mathrm{n}-1) 9=350-17\)
\(\mathrm{n}-1=\frac{333 }{ 9}=37\)
\(\mathrm{n}=38\)
and we know, sum of '\( n \)' terms of an AP if first and last term is given is
\( S_{n}=\frac{n}{2}\left(a+a_{n}\right)\)
\(=\frac{38}{2}[17+350]\)
\(=19 \times 367\)
\(=6973\)

We have; \( \mathrm{n}=22, \mathrm{d}=7 \) and \( \mathrm{a}_{22}=149 \)
where n is number of terms and
d is common difference.
We know;
\( a_{n}=a+(n-1) d \)
\( \Rightarrow 149=\mathrm{a}+(22-1) 7 \)
\(\Rightarrow 149=\mathrm{a}+147\)
\( \Rightarrow \mathrm{a}=149-147\)
\( \Rightarrow \mathrm{a}=2 \)
The sum can be calculated as follows:
\( S=\frac{N}{2}[2 a+(n-1) d]\)
\(=\frac{22}{2}[2+149]\)
\(=11 \times 151\)
\(=1661\)

To Find: \( \mathrm{S}_{51} \)
Given: \( \mathrm{a}_{2}=14, \mathrm{a}_{3}=18 \) and \(n=51\)
We have; \( \mathrm{a}_{2}=14, \mathrm{a}_{3}=18 \) and \( \mathrm{n}=51 \)
We know,
\( d=a_{3}-a_{2}=18-14=4 \)
Therefore, \( \mathrm{d}=4 \)
\( a_{2}-a=4\)
\(14-a=4\)
\(a=10\)
So, first term \( =10 \)
Now, sum can be calculated as follows:
\( S=\frac{N}{2}[2 a+(n-1) d] \)
Where, a and d are first term and common difference respectively.
\(=\frac{51}{2}[2 \times 10+50 \times 4]\)
\(=51 \times\frac{(20+200) }{ 2}\)
\(=51 \times 110\)
\( =5610 \)
\( S_{51}=5610 \)