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Question
In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms?
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Solution
In the given problem, let us take the first term as a and the common difference d
Here, we are given that,
`a_5 = 30` ....(1)
`a_12 = 65` .....(2)
Also, we know
`a_n = a + (n - 1)d`
For the 5th term (n = 5)
`a_5 = a + (5 - 1)d`
30 = a + 4d (Using 1)
a = 30 - 4d ....(3)
Similarly for the 12 th term (n = 12)
`a_12 = a + (12 - 1)d`
65 = a + 11d (Using 2)
a = 65 - 11d....(4)
Substracting (3) from (4) we get
a - a = (65 - 11d)-(30 - 4d)
0 = 65 - 11d - 30 + 4d
0 = 35 - 7d
7d = 35
d = 5
Now, to find a, we substitute the value of d in (4),
a = 30 - 4(5)
a = 30 - 20
a = 10
So for the given A.P d = 5 and a = 10
So to find the sum of first 20 terms of this A.P. we use the following formula for the sum of n terms of an A.P
`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 = 20 we get
`S_20 = 20/2 [2(10) + (20 - 1)(5)]`
= (10)[20 + (19)(5)]
= (10)[20 + 95]
= (10)[115]
= 1150
Therefore, the sum of first 20 terms for the given A.P. is `S_20 = 1150`
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