CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 || NCERT Solutions for Class 10 Maths Chapter 5: Arithmetic Progressions (English Medium)

Here, first term, \( a=121 \) common difference, \( d=a_{2} - a_{1}=117-121=-4\)
Let the first negative term be '\( a_{n} \)'.
Also, we know that, nth term of an AP is given by \( a_{n}=a+(n-1) d \)
We have to find least value of \( n \), such that \( a_{n} < 0 \)
\( \Rightarrow \mathrm{a}+(\mathrm{n}-1) \mathrm{d} < 0 \)
\(\Rightarrow 121+(\mathrm{n}-1)(-4) < 0\)
\(\Rightarrow 121-4(\mathrm{n}-1) < 0\)
\(\Rightarrow 4(\mathrm{n}-1) > 121\)
\(\Rightarrow 4 \mathrm{n}-4 > 121\)
\(\Rightarrow 4 \mathrm{n} > 125\)
\(\Rightarrow \mathrm{n} > 31.25\)
Therefore, n is 32 [least positive integer greater than 31.25 is 32]
Hence, the 32nd term of AP is first negative term. Also, \( a_{32}=a+31 d \) \( =121+31(-4)=121-124=-3 \)

To find: \( \mathrm{S}_{16} \)
Given \( a_{3}+a_{7}=6 \)
\( a_{3} \times a_{7}=8 \)
nth term of an AP is given by the formula \( a_{n}=a+(n-1) d \) where, \( a_{n}= \) nth term \( \mathrm{n}= \) number of term \( \mathrm{d}= \) common difference. So now its given that sum of third and seventh term is 6, thus we need to find 3rd and 7th term first, \( \mathrm{a}_{3}=\mathrm{a}+2 \mathrm{d} \)
\( a_{7}=a+6 d \)
As per question;
\( a_{3}+a_{7}=6 \)
So now,
\(a+2 d+a+6 d=6\)
\(2 a+8 d=6\)
\(a+4 d=3\)
\(a=3-4 d\quad \ldots\ldots\text{(i)}\)
Similarly,
Product of third and seventh term is given as 8. So,
\((a+2 d)(a+6 d)=8\)
\(a^{2}+6 a d+2 a d+12 d^{2}=8\quad \ldots\ldots\text{(ii)}\)
Substituting the value of a in equation (ii), we get;
\((3-4 d)^{2}+8(3-4 d) d+12 d^{2}=8\)
\(9-24 d+16 d^{2}+24 d-32 d^{2}+12 d^{2}=8\)
\(9-4 d^{2}=8\)
\(2 d=1\)
\(d= \pm \frac{1 }{ 2}\)
Using the value of \( d \) in equation (1), we get;
\(a=3-4 d\)
\(a=3-4 \times \frac{1}{2}\)
\(\text {or, } a=3-2=1\)
Sum of first 16 terms is calculated as follows:
\(S=\frac{n}{2}[2 a+(n-1) d]\)
\(S_{16}=\frac{16}{2}\left[2 \times 1+(16-1) \times \frac{1}{2}\right]\)
\(S_{16}=8[2+(\frac{15 }{ 2})]\)
\(=4 \times 19\)
\(\mathrm{S}_{16}=76\)
Thus, sum of first 16 terms of this AP is 76.
Now by taking \( d=-\frac{1 }{ 2 }\), we get, \( a=3-4(-\frac{1 }{ 2}) a=3+2=5 \)
\( S=\frac{16}{2}\left[2 \times 5+(16-1) \frac{-1}{2}\right] \)
\( \mathrm{S}=8[10-\frac{15 }{ 2}] \mathrm{S}=4[20-15] \mathrm{S}=4[5]=20 \)
So, another possible value of sum is 20.