Advertisements
Advertisements
Question
Find the sum of first 51 terms of an A.P. whose 2nd and 3rd terms are 14 and 18 respectively.
Advertisements
Solution
Given, t2 = 14 and t3 = 18
`=>` d = t3 – t2
= 18 – 14
= 4
Now t2 = 14
`=>` a + d = 14
`=>` a = 14 – 4
`=>` a = 10
Sum of n terms of an A.P. = `n/2[2a + (n - 1)d]`
∴ Sum of first 51 terms of an A.P. = `51/2 [2 xx 10 + 50 xx 4]`
= `51/2 [20 + 50 xx 4]`
= `51/2 [20 + 200]`
= `51/2 xx 220`
= 51 × 110
= 5610
APPEARS IN
RELATED QUESTIONS
Check whether -150 is a term of the A.P. 11, 8, 5, 2, ....
Find the sum of the following arithmetic progressions:
41, 36, 31, ... to 12 terms
Show that (a – b)2, (a2 + b2) and (a2 + b2) are in AP.
The sum of the first n terms of an AP is given by Sn = (3n2 – 4n). Find its
- nth term,
- first term and
- common difference.
The sum of the first n terms of an AP is `((5n^2)/2 + (3n)/2)`. Find the nth term and the 20th term of this AP.
In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.
The sum of first 14 terms of an A.P. is 1050 and its 14th term is 140. Find the 20th term.
Which term of the AP 3, 15, 27, 39, ...... will be 120 more than its 21st term?
Find the sum of 12 terms of an A.P. whose nth term is given by an = 3n + 4.
The first term of an AP is –5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.
