Let 3,7,11,15, ... .,403 and 2,5,8,11, ... .,404 be two arithmetic progressions. Then the sum of the common terms in them, is equal to

A.P. : 3,7,11,15, ... , 403 d_1=4 A.P : 2,5,8,11, ... , 404 d_2=3 L.C.M. (d_1, d_2)=12 Common terms =11,23,35 ... 395 \ a=11, d=12, a_n=395 \ a_n=a+(n-1) d => 395 = 11+(n-1)(12) \ => (384 / 12) =(n-1) => n=33 S=(33 / n)(11+395) =(33 / n)(406)=6699

Let 3,7,11,15,.,403 and 2,5,8,11,.,404 be two arithme... | Sequence & Series (Mathematics)

$A.P. : 3,7,11,15, \ldots, 403 d_1=4\\$ A.P $: 2,5,8,11, \ldots, 404$ $d_2=3 \\$ L.C.M. $\left(d_1, d_2\right)=12 \\$ Common terms $=11,23,35 \ldots 395$ $\\ a=11, d=12, a_n=395$ $\\ a_n=a+(n-1) d \Rightarrow 395\\$ = $11+(n-1)(12)$ $\\ \Rightarrow \frac{384}{12}\\$ = $(n-1) \Rightarrow n=33\\$ $S=\frac{33}{n}(11+395) \\=\frac{33}{n}(406)=6699$

Explanation

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