Advertisements
Advertisements
प्रश्न
Find the sum of first 10 terms of the A.P.
4 + 6 + 8 + .............
Advertisements
उत्तर
Here,
First term, a = 4
Common difference, d = 6 – 4 = 2
n = 10
`S = n/2 [2a + (n - 1)d]`
= `10/2 [2(4) + 9(2)]`
= 5[8 + 18]
= 5 × 26
= 130
संबंधित प्रश्न
Find the sum given below:
`7 + 10 1/2 + 14 + ... + 84`
Find the 25th term of the AP \[- 5, \frac{- 5}{2}, 0, \frac{5}{2}, . . .\]
In an A.P. the first term is – 5 and the last term is 45. If the sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?
There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.
There are 25 rows of seats in an auditorium. The first row is of 20 seats, the second of 22 seats, the third of 24 seats, and so on. How many chairs are there in the 21st row ?
Write the value of a30 − a10 for the A.P. 4, 9, 14, 19, ....
Find the sum:
1 + (–2) + (–5) + (–8) + ... + (–236)
Find the sum of all even numbers from 1 to 250.
If the first term of an A.P. is p, second term is q and last term is r, then show that sum of all terms is `(q + r - 2p) xx ((p + r))/(2(q - p))`.
Rohan repays his total loan of ₹ 1,18,000 by paying every month starting with the first installment of ₹ 1,000. If he increases the installment by ₹ 100 every month, what amount will be paid by him in the 30th installment? What amount of loan has he paid after 30th installment?
