CAT 2018 [SLOT 2] Quant Question with Solution 01

N^th term of the series can be written as
$\mathrm{tn}=(4 \mathrm{n}+3)(4 \mathrm{n}+7)$
$=16 \mathrm{n}^{2}+40 \mathrm{n}+21$
$\Sigma \mathrm{tn}=16 \Sigma \mathrm{n}^{2}+40 \Sigma \mathrm{n}+21 \Sigma 1$
$=16 \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+40 \frac{\mathrm{n}(\mathrm{n}+1)}{2}+21 \mathrm{n}$
here n = 23 (7, 11, 15….. 95 is an AP with common different 4 with 23 terms)
$\sum t_{n}=\frac{16 \times 23 \times 24 \times 47}{6}+20 \times 23 \times 24+21 \times 23$
= 80707