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JEE Advanced 2019Paper 1
Exam
JEE Advanced
Year
2019
Paper
Paper 1
Subject
Mathematics
Chapter
Sequences and Series

JEE Advanced 2019 · Paper 1 · Mathematics · Sequences and Series

Let AP(a; d) denote the set of all the terms of an infinite...

Let AP(a;d)AP\left(a; d\right) denote the set of all the terms of an infinite arithmetic progression with first term a and common difference d>0.d > 0 . If AP(1;3)∩AP(2;5)∩AP(3;7)=AP(a;d)AP\left(1; 3\right) \cap AP\left(2; 5\right) \cap AP\left(3; 7\right) = AP\left(a; d\right) then a+da + d equals ____

Correct answer

0

Step-by-step solution

Let T(1,m):mth{T}_{\left(1, m\right)}: {m}^{th} term of 1st{1}^{st} series A(1;3)A\left(1; 3\right)
T(2,n):nth{T}_{\left(2, n\right)}: {n}^{th} term of 2nd{2}^{nd} series A(2;5)A\left(2; 5\right)
T(3,r):rth{T}_{\left(3, r\right)}: {r}^{th} term of 3rd{3}^{rd} series A(3;7)A ( 3; 7 )
For common terms
T(1,m)=T(2,n)=T(3,r){T}_{( 1, m )} = {T}_{( 2, n )} = {T}_{( 3, r )}
1+3(m−1)=2+5(n−1)=3+7(r−1)1 + 3\left(m - 1\right) = 2 + 5\left(n - 1\right) = 3 + 7\left(r - 1\right)
3m−2=5n−3=7r−43m - 2 = 5n - 3 = 7r - 4
⇒m=5n−13andr=5n+17⇒ m = \dfrac{5n - 1}{3}andr = \dfrac{5n + 1}{7}
For ‘m’‘m’ to be a natural number n=2,5,11,…n = 2,5, 11, \ldots
And for ‘r’‘r’ to be a natural number n=4,11,…n = 4,11, \ldots
When n=11,m=18andr=8n = 11, m = 18 and r = 8
⇒⇒ For 1st{1}^{st} common term of three series A(a,d)A ( a, d )
a=1sta = {1}^{st} term of A(1,3)A ( 1,3 )
a=1+(181)3=52a = 1 + ( 181 ) 3 = 52
⇒⇒ Common difference of A(a,d)=LCMA\left(a, d\right) = LCM of common difference of \left(A\left(1; 3\right), A\left(2; 5\right), A\left(3; 7\right)\right)<br/><br />d = LCM ( 3, 5, 7 )<br/><br />d = 105<br/>Hence,<br />Hence,a + d = 52 + 105 = 157
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