Find the 25^{th} common term of the following | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For $S_1$,the first term $a_1 = 1$ and common difference $d_1 = 6 - 1 = 5$.The general term is $T_n = 1 + (n - 1)5 = 5n - 4$.For $S_2$,the first t

Vedclass · Accepted Answer

$491$

Vedclass · Answer

(C) For $S_1$,the first term $a_1 = 1$ and common difference $d_1 = 6 - 1 = 5$.The general term is $T_n = 1 + (n - 1)5 = 5n - 4$.For $S_2$,the first term $a_2 = 3$ and common difference $d_2 = 7 - 3 = 4$.The general term is $T_m = 3 + (m - 1)4 = 4m - 1$.To find common terms,set $5n - 4 = 4m - 1$,which implies $5n = 4m + 3$.The first common term is $11$ (when $n=3, m=3$).The common difference of the new $A.P.$ formed by common terms is $LCM(d_1, d_2) = LCM(5, 4) = 20$.The $k^{th}$ common term is given by $T_k = 11 + (k - 1)20$.For the $25^{th}$ common term $(k = 25)$:$T_{25} = 11 + (25 - 1)20 = 11 + 24 	imes 20 = 11 + 480 = 491$.

Find the $25^{th}$ common term of the following $A.P.'s$
$S_1 = 1, 6, 11, .....$
$S_2 = 3, 7, 11, .....$

Similar Questions

If the $p^{th}$ term of an $A.P.$ is $\frac{1}{q}$ and the $q^{th}$ term is $\frac{1}{p}$,then the sum of its $pq$ terms will be:

The $n^{th}$ term of the series $1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + \dots$ will be

The $3^{rd}$ term of a $G.P.$ is the square of the first term. If the second term is $8,$ determine the $6^{th}$ term.

The product of $n$ positive numbers is unity. Their sum is

If the $9^{th}$ term of an $A.P.$ is zero,then the ratio of its $29^{th}$ term to its $19^{th}$ term is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Find the $25^{th}$ common term of the following $A.P.'s$$S_1 = 1, 6, 11, .....$$S_2 = 3, 7, 11, .....$