Question

The average of 5th and 6th term of an arithmetic progression whose sum is $5n^2–6n$ is equal to

• A. 44
• B. 88
• C. 55
• D. 33

Answer 1

The correct option is A 44

Given: $S_n=5n^2–6n$
Let an be the nth term of the AP. Then,
$a_n=S_n-S_{n-1}$
$=(5n^2-6n)-\left[5\right(n-1{)}^2-6(n-1)]$
$=5n^2-6n-\left[5\right(n^2+1-2n)-6n+6]$
$=5n^2-6n-(5n^2+5-10n-6n+6)$
$=5n^2-6n-(5n^2-16n+11)$
$=10n-11$
Now, $a_5=10\left(5\right)–11=50–11=39$
$a_6=10\left(6\right)–11=60–11=49$
$\therefore \frac{a_5+a_6}{2}=\frac{39+49}{2}=\frac{88}{2}=44$
Hence, the correct answer is option (1).