Advertisements
Advertisements
प्रश्न
If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.
Advertisements
उत्तर
Let a and d be the first term and common difference, respectively of an AP.
∵ Sum of n terms of an AP,
Sn = `n/2 [2a + (n - 1)d]` ...(i)
Now, S6 = 36 ...[Given]
⇒ `6/2[2a + (6 - 1)d]` = 36
⇒ 2a + 5d = 12 ...(ii)
And S16 = 256
⇒ `16/2[2a + (16 - 1)d]` = 256
⇒ 2a + 15d = 32 ...(iii)
On subtracting equation (ii) from equation (iii), we get
10d = 20
⇒ d = 2
From equation (ii),
2a + 5(2) = 12
⇒ 2a = 12 − 10 = 2
⇒ a = 1
∴ S10 = `10/2 [2a + (10 - 1)d]`
= 5[2(1) + 9(2)]
= 5(2 + 18)
= 5 × 20
= 100
Hence, the required sum of first 10 terms is 100.
APPEARS IN
संबंधित प्रश्न
If the nth term of the A.P. 9, 7, 5, ... is same as the nth term of the A.P. 15, 12, 9, ... find n.
Find the 12th term from the end of the following arithmetic progressions:
3, 5, 7, 9, ... 201
Find the sum of all integers between 50 and 500, which are divisible by 7.
Find the sum of the first 11 terms of the A.P : 2, 6, 10, 14, ...
Find the sum 25 + 28 + 31 + ….. + 100
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.
If the 10th term of an AP is 52 and 17th term is 20 more than its 13th term, find the AP
The 19th term of an AP is equal to 3 times its 6th term. If its 9th term is 19, find the AP.
If k,(2k - 1) and (2k - 1) are the three successive terms of an AP, find the value of k.
If the numbers (2n – 1), (3n+2) and (6n -1) are in AP, find the value of n and the numbers
Write the next term for the AP` sqrt( 8), sqrt(18), sqrt(32),.........`
Find the sum of all natural numbers between 200 and 400 which are divisible by 7.
Write an A.P. whose first term is a and common difference is d in the following.
Find the sum of the first 15 terms of each of the following sequences having nth term as xn = 6 − n .
If the sum of first n terms of an A.P. is \[\frac{1}{2}\] (3n2 + 7n), then find its nth term. Hence write its 20th term.
Write the nth term of the \[A . P . \frac{1}{m}, \frac{1 + m}{m}, \frac{1 + 2m}{m}, . . . .\]
The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by \[\frac{l^2 - a^2}{k - (l + a)}\] , then k =
The 9th term of an A.P. is 449 and 449th term is 9. The term which is equal to zero is
The common difference of the A.P.
Q.6
Q.18
Obtain the sum of the first 56 terms of an A.P. whose 18th and 39th terms are 52 and 148 respectively.
For an A.P., If t1 = 1 and tn = 149 then find Sn.
Activitry :- Here t1= 1, tn = 149, Sn = ?
Sn = `"n"/2 (square + square)`
= `"n"/2 xx square`
= `square` n, where n = 75
Find the sum of numbers between 1 to 140, divisible by 4
The first term of an AP of consecutive integers is p2 + 1. The sum of 2p + 1 terms of this AP is ______.
In an A.P., if Sn = 3n2 + 5n and ak = 164, find the value of k.
Find the sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.
Find the value of a25 – a15 for the AP: 6, 9, 12, 15, ………..
Calculate the sum of 35 terms in an AP, whose fourth term is 16 and ninth term is 31.
Assertion (A): a, b, c are in A.P. if and only if 2b = a + c.
Reason (R): The sum of first n odd natural numbers is n2.
