Advertisements
Advertisements
Question
In an AP Given a12 = 37, d = 3, find a and S12.
Advertisements
Solution
Given that, a12 = 37, d = 3
As an = a + (n − 1)d,
a12 = a + (12 − 1)3
37 = a + 11d
37 = a + 11 × 3
37 = a + 33
a = 37 - 33
a = 4
`S_n = n/2 [a+a_n]`
`S_n = 12/2 [4+37]`
`S_n = 6(41)`
`S_n = 246`
Thus, a = 4 and S12 = 246
APPEARS IN
RELATED QUESTIONS
Find the 9th term from the end (towards the first term) of the A.P. 5, 9, 13, ...., 185
The first and the last terms of an AP are 7 and 49 respectively. If sum of all its terms is 420, find its common difference.
Find the sum of n terms of an A.P. whose nth terms is given by an = 5 − 6n.
Find the sum of all even integers between 101 and 999.
In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms?
Find the sum of the first 40 positive integers divisible by 5
Find the value of x for which (x + 2), 2x, ()2x + 3) are three consecutive terms of an AP.
Write the next term for the AP` sqrt( 8), sqrt(18), sqrt(32),.........`
If the sum of first n terms is (3n2 + 5n), find its common difference.
Find an AP whose 4th term is 9 and the sum of its 6th and 13th terms is 40.
The fourth term of an A.P. is 11. The sum of the fifth and seventh terms of the A.P. is 34. Find its common difference.
The sequence −10, −6, −2, 2, ... is ______.
The A.P. in which 4th term is –15 and 9th term is –30. Find the sum of the first 10 numbers.
Find the A.P. whose fourth term is 9 and the sum of its sixth term and thirteenth term is 40.
Mrs. Gupta repays her total loan of Rs. 1,18,000 by paying installments every month. If the installments for the first month is Rs. 1,000 and it increases by Rs. 100 every month, What amount will she pays as the 30th installments of loan? What amount of loan she still has to pay after the 30th installment?
The sum of the first three numbers in an Arithmetic Progression is 18. If the product of the first and the third term is 5 times the common difference, find the three numbers.
If an = 3 – 4n, show that a1, a2, a3,... form an AP. Also find S20.
An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.
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))`.
