Question

The sum of the 3rd term and the 7th terms of an A.P. is 6 and their product is 8. Find the sum of 16 terms.

Answer 1

Let a and d be the first term and common difference of A.P
nth term of A.P., $a_n=a+(n-1)d$
$\therefore a_3=a+(3-1)d=a+2d{a}_7=a+(7-1)d=a+6dGiven,a_3+a_7=6\therefore (a+2d)+(a+6d)=6\Rightarrow 2a+8d=6\Rightarrow a+4d=3\dots \dots \left(1\right)Given,a_3\times a_7=8\therefore (a+2d)+(a+6d)=8\Rightarrow (3-4d+2d)(3-4d+6d)=8\left[Using\right(1\left)\right]\Rightarrow (3-2d)(3+2d)=8\Rightarrow 9-4d^2=8\Rightarrow 4d^2=1\Rightarrow d^2=\frac{1}{4}\Rightarrow d=\pm \frac{1}{2}a=3-4d=3-4\times \frac{1}{2}=3-2=1Whend=-\frac{1}{2},a=3-4d=3-4\times (-\frac{1}{2})=3+2=5$
Whena = 1 and
$d=\frac{1}{2},S_{16}=\frac{16}{2}[2\times 1+(16-1)\times (-\frac{1}{2})]=8(10-\frac{15}{2})=4\times 5=20$
Thus, the sum of first 16 terms of the A.P is 76 or 20.