Advertisements
Advertisements
Question
Find the sum of 28 terms of an A.P. whose nth term is 8n – 5.
Solution
nth term of an A.P. = tn = 8n – 5
Let a be the first term and d be the common difference of this A.P.
Then,
a = t1
= 8 × 1 – 5
= 8 – 5
= 3
t2 = 8 × 2 – 5
= 16 – 5
= 11
∴ d = t2 – t1
= 11 – 3
= 8
The sum of n terms of an A.P. = `S = n/2 [2a + (n - 1)d]`
`=>` Sum of 28 terms of an A.P. = `28/2 [2 xx 3 + 27 xx 8]`
= 14[6 + 216]
= 14 × 222
= 3108
APPEARS IN
RELATED QUESTIONS
In an AP: Given a = 5, d = 3, an = 50, find n and Sn.
If the 10th term of an AP is 52 and 17th term is 20 more than its 13th term, find the AP
Write an A.P. whose first term is a and common difference is d in the following.
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
The common difference of the A.P.
Show that a1, a2, a3, … form an A.P. where an is defined as an = 3 + 4n. Also find the sum of first 15 terms.
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`.
What is the sum of an odd numbers between 1 to 50?
The sum of first five multiples of 3 is ______.
Find the sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.
