Question

The sum of all terms of the arithmetic progression having ten terms except for the first tens, is 99, and except for the sixth term, is 89. Find the third term of the progression if the sum of the first and the fifth term is equal to 10.

• A. 15
• B. 5
• C. 8
• D. 10

Answer 1

Answer: (B)

5

Given:

$S_{10}=99+T_1..........\left(i\right)S_{10}=89+T_6..........\left(ii\right)$

where $S_{10}$ is the sum of $10$ terms of the A.P. and $T_1,T_6$ are the first and sixth term respectively.

Say $a$ and $d$ are the first term and common difference of the A.P. respectively.

$\therefore S_{10}=5\{2a+9d\};T_1=a;T_6=a+5d........\left(iii\right)\therefore 5\{2a+9d\}=a+\mathrm{99........}\left(iv\right)5\{2a+9d\}=a+89+5d........\left(v\right)$

Subtracting (iv) and (v), we get,

$10-5d=0=>d=\mathrm{2........}\left(vi\right)$

Also given that

$T_1+T_5=10=>a+a+4d=10=>2a+4\times 2=10=>2a=2=>a=1$

$\therefore T_3=a+2d=1+2\times 2=5$