Advertisements
Advertisements
Question
Find the sum of 28 terms of an A.P. whose nth term is 8n – 5.
Advertisements
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
The sum of the first p, q, r terms of an A.P. are a, b, c respectively. Show that `\frac { a }{ p } (q – r) + \frac { b }{ q } (r – p) + \frac { c }{ r } (p – q) = 0`
In an AP given a = 8, an = 62, Sn = 210, find n and d.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Find the sum of all natural numbers between 1 and 100, which are divisible by 3.
The first term of an AP is p and its common difference is q. Find its 10th term.
Find the sum of first n terms of an AP whose nth term is (5 – 6n). Hence, find the sum of its first 20 terms.
If first term of an A.P. is a, second term is b and last term is c, then show that sum of all terms is \[\frac{\left( a + c \right) \left( b + c - 2a \right)}{2\left( b - a \right)}\].
x is nth term of the given A.P. an = x find x .
The sum of the first 15 multiples of 8 is ______.
Find the sum of first 20 terms of an A.P. whose nth term is given as an = 5 – 2n.
