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
Answers
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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