Advertisements
Advertisements
प्रश्न
The nth term of an Arithmetic Progression (A.P.) is given by the relation Tn = 6(7 – n)..
Find:
- its first term and common difference
- sum of its first 25 terms
Advertisements
उत्तर
Given, Tn = 6(7 – n)
a. For first term, put n = 1
Then, a1 = 6(7 – 1)
= 6 × 6
= 36
For second term, put n = 2
Then a2 = 6(7 – 2)
= 6 × 5
= 30
Then, common difference
∴ d = a2 – a1
= 30 – 36
= – 6
Hence, first term is 36 and common difference is - 6.
b. `S_n = n/2[2a + (n - 1)d]`
`S_25 = 25/2[2 xx 36 + (25 - 1)(-6)]`
= `25/2[72 - 144]`
= `25/2 xx (-72)`
S25 = – 900
APPEARS IN
संबंधित प्रश्न
Find the sum of the following APs.
−37, −33, −29, …, to 12 terms.
The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceed the second term by 6, find three terms.
If the 10th term of an AP is 52 and 17th term is 20 more than its 13th term, find the AP
If `4/5 `, a, 2 are in AP, find the value of a.
If the sum of first p terms of an AP is 2 (ap2 + bp), find its common difference.
Find the sum of the first 15 terms of each of the following sequences having nth term as xn = 6 − n .
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 =
Q.16
Find the sum of all odd numbers between 351 and 373.
The sum of 40 terms of the A.P. 7 + 10 + 13 + 16 + .......... is ______.
