Question

if sum of 5 consecutive terms of an a.p is 115. then find the third term. ​

Answer 1

$\bold\red{\underline{\underline{Answer:}}}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{Let \ five \ consecutive \ terms \ of \ A.P. \ be}$

$\bold\orange{a, \ a+d, \ a+2d, \ a+3d, \ a+4d}$

$\bold\green{According \ to \ the \ given \ condition}$

$\bold{(a)+(a+d)+(a+2d)+(a+3d)+(a+4d)=115}$

$\bold{4a+10d=115}$

$\bold{2(2a+5d)=115....(1)}$

$\bold{Sn=n×2^{-1}[2a+(n-1)d]}$

$\bold{S5=5×2^{-1}[2a+4d]....(2)}$

2(2a+5d)=5×2^-1[2a+4d]....from(1) & (2)

$\bold{5a+10d=4a+10d}$

$\bold{a=0}$

$\bold{Substitudind \ a=0 \ in \ equation(1)}$

$\bold{2(0+5d)=115}$

$\bold{10d=115}$

$\bold{d=11.5}$

$\bold{tn=a+(n-1)d=115....(formula)}$

$\bold{t3=0+(3-1)11.5}$

$\bold{t3=2×11.5}$

$\bold{t3=23}$

$\bold\green{Third \ term \ is \ 23.}$