JEE Main 2019 — Sequences and Series Question with Solution

Two-digit numbers of the form $7\lambda +2$ are $16, 23, 30,..., 93$

These number are forming an arithmetic progression with first term $a=16,$ common difference $d=23-16=7$ and last term $l=93$

And, we know that the last term of an arithmetic progression is given by $l=a+\left(n-1\right)d$

$\Rightarrow 93=16+\left(n-1\right)7$

$\Rightarrow 77=\left(n-1\right)7$

$\Rightarrow n=12$

Two-digit numbers of the form $7\lambda +5$ are $12, 19, 26,..., 96$

Again, using the last term of $A.P.,$ we get $96=12+\left(n_1-1\right)7$

$\Rightarrow 84=\left(n_1-1\right)7$

$\Rightarrow n_1=13$

Now, the sum of $n$ terms of an $A.P.$ is $S_n=\frac{n}{2}\left(a+l\right),$ we get the sum of all the above numbers as

$=\frac{12}{2}\left(16+93\right)+\frac{13}{2}\left(12+96\right)$

$=654+702=1356.$