Advertisements
Advertisements
Question
Determine the A.P. whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.
Advertisements
Solution
We have given a3 = 16
a + (3 − 1)d = 16
a + 2d = 16(1)
a7 − a5 = 12
[a + (7 − 1)d] − [a + (5 − 1)d] = 12
(a + 6d) − (a + 4d) = 12
2d = 12
d = 6
From equation (1), we obtain
a + 2(6) = 16
a + 12 = 16
a = 4
Therefore, A.P. will be 4, 10, 16, 22, ...
APPEARS IN
RELATED QUESTIONS
In an AP given a = 8, an = 62, Sn = 210, find n and d.
Find how many integers between 200 and 500 are divisible by 8.
Find the sum of the following arithmetic progressions:
3, 9/2, 6, 15/2, ... to 25 terms
If 12th term of an A.P. is –13 and the sum of the first four terms is 24, what is the sum of first 10 terms?
In an A.P., if the first term is 22, the common difference is −4 and the sum to n terms is 64, find n.
Choose the correct alternative answer for the following question.
For an given A.P. a = 3.5, d = 0, n = 101, then tn = ....
The sum of the first n terms of an A.P. is 3n2 + 6n. Find the nth term of this A.P.
If 18, a, b, −3 are in A.P., the a + b =
The 11th term and the 21st term of an A.P are 16 and 29 respectively, then find the first term, common difference and the 34th term.
If the first term of an A.P. is 5, the last term is 15 and the sum of first n terms is 30, then find the value of n.
