Advertisements
Advertisements
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.
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.
APPEARS IN
RELATED QUESTIONS
How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Find the sum of all integers between 50 and 500, which are divisible by 7.
Find the sum of all even integers between 101 and 999.
Find the sum of the first 40 positive integers divisible by 5
Find the sum of the first 25 terms of an A.P. whose nth term is given by an = 2 − 3n.
Which term of the AP 3,8, 13,18,…. Will be 55 more than its 20th term?
Simplify `sqrt(50)`
The sum of third and seventh term of an A. P. is 6 and their product is 8. Find the first term and the common difference of the A. P.
If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, ..., is 116. Find the last term.
Sum of 1 to n natural number is 45, then find the value of n.