Question

if tn respresents nth term of an AP t2 + t5 - t3=10 and t2 + t9 = 17, find it's first term and common difference.​

Answer 1

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

$\sf{First \ term \ is \ 13 \ and \ common \ difference}$

$\sf{is \ -1.}$

$\sf\orange{Given:}$

$\sf{In \ an \ AP}$

$\sf{\implies{t2+t5-t3=10}}$

$\sf{\implies{t2+t9=17}}$

$\sf\pink{To \ find:}$

$\sf{Find \ the \ first \ term \ and \ common \ difference.}$

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

$\boxed{\sf{tn=a+(n-1)d}}$

$\sf{According \ to \ the \ first \ condition.}$

$\sf{t2+t5-t3=10}$

$\sf{\therefore{(a+d)+(a+4d)-(a+2d)=10}}$

$\sf{\therefore{a+3d=10...(1)}}$

$\sf{According \ to \ the \ second \ condition.}$

$\sf{t2+t9=17}$

$\sf{\therefore{(a+d)+(a+8d)=17}}$

$\sf{\therefore{2a+9d=17...(2)}}$

$\sf{Multiply \ equation \ (1) \ by \ (3)}$

$\sf{3a+9d=30...(3)}$

$\sf{Subtract \ equation \ (2) \ from \ equation \ (3)}$

$\sf{3a+9d=30}$

$\sf{-}$

$\sf{2a+9d=17}$

__________________

$\boxed{\sf{\therefore{a=13}}}$

$\sf{Substitute \ a=13 \ in \ equation \ (1)}$

$\sf{13+3d=10}$

$\sf{3d=10-13}$

$\sf{3d=-3}$

$\sf{d=\frac{-3}{3}}$

$\boxed{\sf{d=-1}}$

$\sf\purple{\tt{\therefore{First \ term \ is \ 13 \ and \ common \ difference}}}$

$\sf\purple{\tt{is \ -1.}}$