Question

the sum of the third and the seventh terms of an AP js 6 and their product is 8. Find the sum of the first sixteen terms of the AP​

Answer 1

$\sf\large\underline{Given:}$

• $\rm{The\:sum\:of\:3rd\:and\:7th\: term=6}$
• $\rm{The\: product\:of\:the\:term=8}$

$\sf\large\underline{To\: Find:}$

• $\rm{The\:sum\:of\:1st\:16th\: terms=?}$

$\sf\large\underline{Solution:}$

• At first find the value of first term and common difference of AP. Than calculate the sum of 1st 16th terms]

$\rm{\implies Let,\: first\:term=a\: common\: difference=d}$$\sf\large\underline{Formula\: used:}$

$\rm{\implies T_{n}=a+(n-1)d}$

$\rm{\implies The\:sum\:of\:3rd\:7th\: terms=6}$

$\rm{\implies a+2d+a+6d=6}$

$\rm{\implies 2a+8d=6}$

• Dividing by 2 on both sides]

$\rm{\implies a+4d=3-----(i)}$

$\rm{\implies a=3-4d}$

• Now, calculate product of the terms]

$\rm{\implies (a+2d)(a+6d)=8}$

$\rm{\implies a^2+8ad+12d^2-8=0----(ii)}$

• Putting the value of a=3-4d here]

$\rm{\implies (3-4d)^2+8(3-4d)d+12d^2-8=0}$

$\rm{\implies 9-24d+16d^2+24d-32d^2-8=0}$

$\rm{\implies -16d^2+1=0}$

$\rm{\implies 16d^2=1}$

$\rm{\implies d=\dfrac{1}{4}}$

• Now putting the value of d=1/4 in eq 1]

$\rm{\implies a+4d=3}$

$\rm{\implies a+4\times\dfrac{1}{4}=3}$

$\rm{\implies a+1=3}$

$\rm{\implies a=3-1}$

$\rm{\implies a=2}$

• Now, calculate the sum of 1st 16th term by using formula]

$\rm{\implies S_{n}=\dfrac{n}{2}[2a+(n-1)d]}$

$\rm{\implies S_{16}=\dfrac{16}{2}[2\times\:2+(16-1)\dfrac{1}{4}]}$

$\rm{\implies S_{16}=\dfrac{16}{2}[4+\dfrac{15}{4}]}$

$\rm{\implies S_{16}=\dfrac{16}{2}[\dfrac{16+15}{4}]}$

$\rm{\implies S_{16}=\dfrac{16}{2}\times\dfrac{31}{4}}$

$\rm{\implies S_{16}=31\times 2}$

$\rm{\implies S_{16}=62}$

$\sf\large{Hence,}$

$\rm{\implies The\:sum\:of\:1st\:16th\: terms=62}$