Complicated sum of terms in an AP

Algebra Level pending

Consider an arithmetic progression (AP) whose first term is 1000 and common difference is 242. Find the sum of 10000 terms of this AP with 64 terms between two successive terms which are to be included in the sum, starting from the 3520th term.

For example, in the AP: 1,3,5,7,9,11,13,15,17,19.... The terms 1 and 5 have one term, 3, between them. Here, the sum of 3 terms with 2 terms between two successive terms which are to be included in the sum, starting from the 4th term would mean the following: 7+13+19 because the sum contains a total of three terms as it has been said above, two successive terms that are to be included in the sum have two terms between them (for example, 7 and 13 have two terms, 9 and 11 between them) and the sum starts with the 4th term in the AP.

The answer is 794947330000.

1 solution

Tapas Mazumdar
Apr 29, 2017

The $3520^{\text{th}}$ term of this AP is

$a_{3520} = 1000 + 3519 \times 242 = 852598$

Note that we need to discard $64$ terms after each successive term starting from $852598$ . Thus the next term to be considered in our sum would be the $65^{\text{th}}$ term after $852598$ which is

$a_{3585} = a_{3520} + 65 \times 242 = 852598 + 65 \times 242 = 868328$

We construct a new AP with only those terms of the original AP that fulfill the condition of being in gaps with $64$ terms in between them, this new AP has a common difference

$d = 868328 - 852598 = 15730$

And so the required sum is

$S = \dfrac{10000}{2} \left(2 \times 852598 + 9999 \times 15730 \right) = \boxed{794947330000}$

Complicated sum of terms in an AP

The answer is 794947330000.

1 solution

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