Advertisements
Advertisements
Question
Find the sum of first 51 terms of an A.P. whose 2nd and 3rd terms are 14 and 18 respectively.
Advertisements
Solution
Given, t2 = 14 and t3 = 18
`=>` d = t3 – t2
= 18 – 14
= 4
Now t2 = 14
`=>` a + d = 14
`=>` a = 14 – 4
`=>` a = 10
Sum of n terms of an A.P. = `n/2[2a + (n - 1)d]`
∴ Sum of first 51 terms of an A.P. = `51/2 [2 xx 10 + 50 xx 4]`
= `51/2 [20 + 50 xx 4]`
= `51/2 [20 + 200]`
= `51/2 xx 220`
= 51 × 110
= 5610
RELATED QUESTIONS
The sum of three numbers in A.P. is –3, and their product is 8. Find the numbers
Find the sum of the following APs.
`1/15, 1/12, 1/10`, ......, to 11 terms.
Find the sum of the first 40 positive integers divisible by 5
Find the sum of two middle most terms of the AP `-4/3, -1 (-2)/3,..., 4 1/3.`
Draw a triangle PQR in which QR = 6 cm, PQ = 5 cm and times the corresponding sides of ΔPQR?
The 9th term of an A.P. is equal to 6 times its second term. If its 5th term is 22, find the A.P.
The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its
The common difference of the A.P. \[\frac{1}{2b}, \frac{1 - 6b}{2b}, \frac{1 - 12b}{2b}, . . .\] is
Kanika was given her pocket money on Jan 1st, 2008. She puts Rs 1 on Day 1, Rs 2 on Day 2, Rs 3 on Day 3, and continued doing so till the end of the month, from this money into her piggy bank. She also spent Rs 204 of her pocket money, and found that at the end of the month she still had Rs 100 with her. How much was her pocket money for the month?
The sum of 41 terms of an A.P. with middle term 40 is ______.
