Question

QUESTION-13
If S1 = 3, 7, 11, 15,....upto 125
terms and S2 = 4, 7, 10, 13,
16........upto 125 terms, then
how many terms are there in S1
that are there in S2?
29

Answer 1

Given : S1 = 3, 7, 11, 15, .upto 125 terms and S2 = 4, 7, 10, 13, 16 . upto 125 terms,

To Find : how many terms are there in S1 that are there in S2

Solution:

S1 = 3, 7, 11, 15, ..upto 125 terms

a = 3

d = 4

Last term  = 3 + (125 - 1)4 = 499

S1 =4, 7, 10, 13,16. upto 125 terms

a = 4

d = 3  

Last term  = 4 + (125 - 1)3 = 376

1st common term = 7

After that   3 * 4 = 12      will be common difference

a = 7

d = 12

Last term ≤ 376

7 + (n - 1) 12   ≤ 376

=>  (n - 1) 12   ≤ 369

=> (n - 1)    ≤ 30

=>  n ≤ 31

31 terms are there in S1 that are there in S2

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