The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.
If the ratio of the sum of first n terms of two A.P’s is (7n +1): (4n + 27), find the ratio of their mth terms.
The houses in a row numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses preceding the house numbered X is equal to the sum of the numbers of houses following X.
In an AP, if S5 + S7 = 167 and S10 = 235, then find the AP, Where Sn denotes the sum of its first n terms.
The 14th term of an AP is twice its 8th term. If its 6th terms is -8, then find the sum of its first 20 terms.
Let a and d be the first term and the common difference of the AP, respectively.
∴ nth term of the AP, an = a + (n-1)d
So,
a14 = a+ (14-1)d = a + 13d
a8 = a + (8 - 1)d = a + 7d
a6 = a+ (6 - 1)d = a + 5d
According to the question,
a14 = 2a8
⇒ a + 13d = 2 (a + 7d)
⇒ a + d = 0 .... (i)
Also,
a6 = a + 5d = - 8 ... (ii)
solving (i) and (ii), we get
a = 2 and d = -2
∴ S20 = 20/2 [2 x 2 + (20 - 1)(-2)]
= - 340
Hence, the sum of the first 20 terms is -340.
Find the 60th term of the AP 8, 10,12, ..., if it has a total of 60 terms and hence find the sum of its last 10 terms.