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The sum of 4th and 13th term of an AP is 35. What is the sum

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The sum of 4th and 13th term of an AP is 35. What is the sum [#permalink]

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New post 31 Jan 2012, 17:53
The sum of 4th and 13th term of an AP is 35. What is the sum of the first 15 terms of the AP ?

Could someone please provide a solution to this problem ?
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Re: The sum of 4th and 13th term of an AP is 35. What is the sum [#permalink]

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New post 31 Jan 2012, 19:59
We can find the answer for the sum of the first 16 terms of the AP.

The 4th term of an AP = a + 3d [a= first term and d=common difference]
The 13th term of the AP = a + 12d
=> 2a + 15d = 35

Sum of the first 16 terms of the AP = (16/2) [2a + 15d]
= 8 * 35
= 280
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Last edited by GyanOne on 03 Feb 2012, 00:55, edited 1 time in total.
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Re: The sum of 4th and 13th term of an AP is 35. What is the sum [#permalink]

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New post 01 Feb 2012, 01:03
Thanks for the solution
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Re: The sum of 4th and 13th term of an AP is 35. What is the sum [#permalink]

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New post 02 Feb 2012, 10:29
Hi GyanOne,
Is not the sum of the first n terms of an AP series given by :

S(n) = (n/2) [2a + (n-1)d]

so S(15) = (15/2) [2a +14d] ?
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Re: The sum of 4th and 13th term of an AP is 35. What is the sum [#permalink]

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New post 03 Feb 2012, 00:56
radhagovindaradjou,

We can't sum for 15 terms here, but can for 16. The expression you have put down is correct, except we will not be able to find a numerical value for the expression. We can sum to 16 terms though, and that is what I have done.
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Re: The sum of 4th and 13th term of an AP is 35. What is the sum   [#permalink] 03 Feb 2012, 00:56
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The sum of 4th and 13th term of an AP is 35. What is the sum

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