Advertisements
Advertisements
Question
In an A.P. the first term is 25, nth term is –17 and the sum of n terms is 132. Find n and the common difference.
Advertisements
Solution
First term a = 25
nth term = –17
`=>` Last term l = –17
Sum of n terms = 132
`=> n/2 [a + l] = 132`
`=>` n(25 – 17) = 264
`=>` n × 8 = 264
`=>` n = 33
Now, l = –17
`=>` a + (n – 1)d = –17
`=>` 25 + 32d = –17
`=>` 32d = – 42
`=> d = -42/32`
`=> d = -21/16`
APPEARS IN
RELATED QUESTIONS
In an AP, given a = 2, d = 8, and Sn = 90, find n and an.
Find the sum of the first 11 terms of the A.P : 2, 6, 10, 14, ...
Find the sum of first 51 terms of an A.P. whose 2nd and 3rd terms are 14 and 18 respectively.
Is –150 a term of the AP 11, 8, 5, 2, ...?
Find the A.P. whose fourth term is 9 and the sum of its sixth term and thirteenth term is 40.
In an A.P., the first term is 22, nth term is −11 and the sum to first n terms is 66. Find n and d, the common difference
Find the sum of natural numbers between 1 to 140, which are divisible by 4.
Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136
Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
`square` = (n – 1) × 4
n = `square`
Now,
Sn = `"n"/2["a" + "t"_"n"]`
Sn = 17 × `square`
Sn = `square`
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
The sum of first n terms of the series a, 3a, 5a, …….. is ______.
The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235, find the sum of its first twenty terms.
An Arithmetic Progression (A.P.) has 3 as its first term. The sum of the first 8 terms is twice the sum of the first 5 terms. Find the common difference of the A.P.
