Question

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

Answer 1

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

Answer 2

✯ 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.

✰ BE BRAINLY ✰

THANKS!!