JEE Main 2019 — Mathematics Algebra

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

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

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

$$\Rightarrow 93=16+(n-1)7$$⇒93=16+(n−1)7

$$\Rightarrow 77=(n-1)7$$⇒77=(n−1)7

$$\Rightarrow n=12$$⇒n=12

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

Again, using the last term of $A.P.,$A.P., we get $96=12+({n}_{1}-1)7$96=12+(n1​−1)7

$$\Rightarrow 84=({n}_{1}-1)7$$⇒84=(n1​−1)7

$$\Rightarrow {n}_{1}=13$$⇒n1​=13

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

$$=\frac{12}{2}(16+93)+\frac{13}{2}(12+96)$$=212​(16+93)+213​(12+96)

$$=654+702=1356.$$=654+702=1356.