Advertisements
Advertisements
प्रश्न
The first term of an A.P. is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.
Advertisements
उत्तर
First term a = 5
Last term l = 45
Sum of terms = 1000
Let there be n terms in this A.P.
Now, sum of first n terms = `n/2 [a + l]`
`=> 1000 = n/2 [5 + 45]`
`=>` 2000 = n × 50
`=>` n = 40
l = a + (n – 1)d
`=>` 45 = 5 + (40 – 1)d
`=>` 40 = 39d
`=> d = 40/39`
Hence, number of terms are 40 and common difference is `40/39`.
APPEARS IN
संबंधित प्रश्न
A small terrace at a football field comprises 15 steps, each of which is 50 m long and built of solid concrete. Each step has a rise of `1/4` m and a tread of `1/2` m (See figure). Calculate the total volume of concrete required to build the terrace.
[Hint: Volume of concrete required to build the first step = `1/4 xx 1/2 xx 50 m^3`]
Find the sum to n term of the A.P. 5, 2, −1, −4, −7, ...,
Find the 8th term from the end of the AP 7, 10, 13, ..., 184.
Determine the nth term of the AP whose 7th term is –1 and 16th term is 17.
Write an A.P. whose first term is a and common difference is d in the following.
a = –19, d = –4
Find the sum of the first 15 terms of each of the following sequences having nth term as xn = 6 − n .
Let there be an A.P. with first term 'a', common difference 'd'. If an denotes in nth term and Sn the sum of first n terms, find.
Find the sum of all members from 50 to 250 which divisible by 6 and find t13.
Find the sum:
1 + (–2) + (–5) + (–8) + ... + (–236)
Jaspal Singh repays his total loan of Rs. 118000 by paying every month starting with the first instalment of Rs. 1000. If he increases the instalment by Rs. 100 every month, what amount will be paid by him in the 30th instalment? What amount of loan does he still have to pay after the 30th instalment?
