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
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 20 terms of the A.P. 1, 4, 7, 10, ……
Find the sum of the following APs:
2, 7, 12, ..., to 10 terms.
Find the sum of the first 15 terms of each of the following sequences having the nth term as
yn = 9 − 5n
Find the sum of the first 22 terms of the A.P. : 8, 3, –2, ………
How many terms of the AP 21, 18, 15, … must be added to get the sum 0?
The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find its common difference.
The sum of first n terms of an A.P is 5n2 + 3n. If its mth terms is 168, find the value of m. Also, find the 20th term of this A.P.
Write the nth term of the \[A . P . \frac{1}{m}, \frac{1 + m}{m}, \frac{1 + 2m}{m}, . . . .\]
The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is
Find the sum:
1 + (–2) + (–5) + (–8) + ... + (–236)