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Question
The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is
Options
5
10
12
14
20
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Solution
In the given problem, we have an A.P. 3,7,11,15,....
Here, we need to find the number of terms n such that the sum of n terms is 406.
So here, we will use the formula,
`S_n = n/2[2a + (n-1)d ]`
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
The first term (a) = 3
The sum of n terms (Sn) = 406
Common difference of the A.P. (d) = `a_2 - a_1`
= 7 - 3
= 4
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,
`406 = n/2 [ 2(3) + (n-1) (4) ] `
`406 = (n/2) [ 6 +(4n- 4)]`
`406 = (n/2) [ 2 + 4n]`
`406 = n + 2n^2`
So, we get the following quadratic equation,
`2n^2 + n - 406 = 0`
On solving by splitting the middle term, we get,
`2n^2 - 28n + 29n - 406 = 0`
`2n ( n- 14) - 29 (n-14)= 0`
`(2n - 29 ) ( n- 14) = 0`
Further,
2n - 29 = 0
`n = 29/2`
Or,
n - 14 = 0
n = 14
Since, the number of terms cannot be a fraction, the number of terms (n) is n = 14
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