If 3 + 5 + 7 + ...+ n terms / 5 + 8 + 11 + ...+ 10 terms =7 then the value of n is (a) 35 (b) 36 (c) 37 (d) 40
Answers
Answer:
35
Step-by-step explanation:
Sum of first n term of AP = (n/2) [2a + (n - 1)d]
Solving the numerator:
3 + 5 + 7 + ... n terms =
= (n/2) [2(3) + (n - 1)2]
= n(n + 2)
Solving the denominator:
5 + 8 + 11 + ... upto 10 terms =
= (10/2) [2(5) + (10 - 1)3]
= 185
Hence, the question is:
⇒ n(n + 2)/185 = 7
⇒ n² + 2n = 1295
⇒ n² + 37n - 35n - 1295 = 0
⇒ n(n + 37) - 35(n + 37) = 0
⇒ (n - 35)(n + 37) = 0
Hence, n = 35
✯ SOLUTION ✯
Given (3+5+7+… n tems)/(5+8+11+…10 terms) = 7
Consider (3+5+7+… n tems)
This is an AP with a = 3 and d = 2
Sum of n terms = (n/2)(2a+(n-1)d)
= (n/2)(6+(n-1)2)
= (n/2)(4+2n)
= n(2+n)
Consider (5+8+11+…10 terms)
This is an AP with a = 5 and d = 3.
S10 = (10/2)(10+9(3))
= 5×37
(3+5+7+… n tems)/(5+8+11+…10 terms) = 7
n(2+n)/5×37 = 7
n(2+n) = 7×5×37
n(n+2) = 35×37
So n = 35
Hence option (1) is the answer.