Short Answer TypeThe 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.
We have given that
4th term of an A.P.= a4 = 0
∴ a + (4 – 1)d = 0
∴ a + 3d = 0
∴ a = –3d ….(1)
25th term of an A.P. = a25
= a + (25 – 1)d
= –3d + 24d ….[From the equation (1)]
= 21d
3 times 11th term of an A.P. = 3a11
= 3[a + (11 – 1)d]
= 3[a + 10d]
= 3[–3d + 10d]
= 3 × 7d
= 21d
∴ a25 = 3a11
i.e., 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.
Long Answer TypeFind 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.
Short Answer Type
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