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Question
If the 12th term of an A.P. is −13 and the sum of the first four terms is 24, what is the sum of first 10 terms?
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Solution
In the given problem, we need to find the sum of first 10 terms of an A.P. Let us take the first term a and the common difference as d
Here, we are given that,
`a_12 = -13`
`S_4 = 24`
Also, we know
`a_n= a + (n - 1)d`
For the 12th term (n = 12)
`a_12 = a + (12 - 1)d`
`-13 = a + 11d`
a= -13 - 11d ......(1)
So, as we know the formula for the sum of n terms of an A.P. is given by,
`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
So, using the formula for n = 4, we get,
`S_4 = 4/2 [2(a) + (4 - 1)(d)]`
24 = (2)[2a + (3)(d)]
24 = 4a +6d
4a = 24 - 6d
` a= 6 - 6/4 d` ....(2)
Subtracting (1) from (2), we get,
`a - a = (6 - 6/4 d) - (-13 - 11)d`
`0 = 6 - 6/4 d + 13 + 11d`
`0 = 19 + 11d - 6/4 d`
`0 =19+ (44d - 6d)/4`
On further simplifying for d, we get,
`0 = 19 + (38d)/4`
`-19=19/2 d`
`d= (-19(2))/2`
d= -2
Now, to find a, we substitute the value of d in (1),
a = -13 -11(-2)
a = -13 + 22
a = 9
Now using the formula for the sum of n terms of an A.P. for n = 10 we get
`S_10 = 10/2 [2(9) + (10 - 1)(-2)]`
= (5)[18 + (9)(-2)]
= (5)(18 - 18)
=(5)(0)
= 0
Therefore the sum of first 10 terms for the given A.P is `S_10 = 0`
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