Find the sum to 200 terms of the series 1 + 4 + 6 + 5 + 11 + 6 +…(a)

1.	Find the sum to 200 terms of the series 1 + 4 + 6 + 5 + 11 + 6 +…(a) 30,200 (b) 29,800 (c) 30,200 (d) None of these
Answer» Spot that the above series is a combination of two A.P.s. The 1st A.P. is (1 + 6 + 11 +…) and the 2nd A.P. is (4 + 5 + 6 +…) Since the terms of the two series alternate, S = (1 + 6 + 11 +… to 100 terms) + (4 + 5 + 6 + … to 100 terms) = 100[2 x 1 +99 x 5]/2+ 100[2 x 4+99 x 1]/2 → (Using the formula for the sum of an AP = 50[497 + 107] = 50[604] = 30200 Alternatively, we can treat every two consecutive terms as one. So we will have a total of 100 terms of the nature: (1 + 4) + (6 + 5) + (11 + 6) …→ 5, 11, 17… Now, a = 5, d = 6 and n = 100 Hence the sum of the given series is S = 100/2 x [ 2 x 5 +99 x 6] = 50 [ 604] = 30200

Find the sum to 200 terms of the series 1 + 4 + 6 + 5 + 11 + 6 +…(a) 30,200 (b) 29,800 (c) 30,200 (d) None of these

Answer»

Spot that the above series is a combination of two A.P.s.

The 1st A.P. is (1 + 6 + 11 +…) and the 2nd A.P. is (4 + 5 + 6 +…)

Since the terms of the two series alternate,

S = (1 + 6 + 11 +… to 100 terms) + (4 + 5 + 6 + … to 100 terms)

= 100[2 x 1 +99 x 5]/2+ 100[2 x 4+99 x 1]/2

→ (Using the formula for the sum of an AP

= 50[497 + 107] = 50[604] = 30200

Alternatively, we can treat every two consecutive terms as one.

So we will have a total of 100 terms of the nature:

(1 + 4) + (6 + 5) + (11 + 6) …→ 5, 11, 17…

Now, a = 5, d = 6 and n = 100

Hence the sum of the given series is

S = 100/2 x [ 2 x 5 +99 x 6]

= 50 [ 604] = 30200

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