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Question
If the second term and the fourth term of an A.P. are 12 and 20 respectively, then find the sum of first 25 terms:
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Solution
For an A.P. t2 = 12 and t4 = 20
To find : S25 = ?
∴ tn = a+(n-1)d
∴ t2 = a+(2-1)d
∴ 12 = a + d .....eq(1)
∴ t4 = a + (4 - 1)d
∴ 20 = a + 3d ....eq(2)
Substracting eq(i) from eq(ii)
a + 3d = 20
`(a + d = 12)/(2d = 8)`
`"d" = 8/2`
∴ d = 4
Substituting d = 4 in eq (i)
a + d = 12
∴ a + 4 = 12
∴ a = 12 - 4
∴ a = 8
`"S"_"n" = "n"/2 ["2a" + ("n" - 1)"d"]`
`therefore "S"_25 = 25/2 [2(8) + (25 - 1)(4)]`
`= 25/2 [16 + 24(4)]`
`= 25/2[16 + 96]`
`= 25/2 xx 112`
= 1400
The sum of first 25 terms is 1400.
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