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Question
If the last term of an A.P. of 30 terms is 119 and the 8th term from the end (towards the first term) is 91, then find the common difference of the A.P. Hence, find the sum of all the terms of the A.P.
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Solution
Given, last term, l = 119
No. of terms in A.P. = 30
8th term from the end = 91
Let d be a common difference and assume that the first term of A.P. is 119 (from the end)
Since the nth term of AP is
an = l + (n – 1)d
∴ a8 = 119 + (8 – 1)d
⇒ 91 = 119 + 7d
⇒ 7d = 91 – 119
⇒ 7d = –28
⇒ d = –4
Now, this common difference is from the end of A.P.
So, the common difference from the beginning = –d
= –(–4) = 4
Thus, a common difference of the A.P. is 4.
Now, using the formula
l = a + (n – 1)d
⇒ 119 = a + (30 – 1)4
⇒ 119 = a + 29 × 4
⇒ 119 = a + 116
⇒ a = 119 – 116
⇒ a = 3
Hence, using the formula for the sum of n terms of an A.P.
i.e., Sn = `"n"/2[2"a" + ("n" - 1)"d"]`
S30 = `30/2[2 xx 3 + (30 - 1) xx 4]`
= 15 (6 + 29 × 4)
= 15 (6 + 116)
= 15 × 122
= 1830
Therefore, the sum of 30 terms of an A.P. is 1830.
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