Short Answer TypeLet the common difference of both the
A.P.’s = d
and First term of Ist A.P. = a
also first term of 2nd A .P. = b
Then, 200th term of 1st A .P. = a + 99d
and 100th term of IInd A .P. = b + 99d
Difference between their 100th term = 100
(given)
⇒ (a + 99d) – (b + 99d) = 100
⇒ a – b = 100 ...(i)
Now, a1000 of Ist A.P. = a + 999d
and a1000 of Ilnd A .P. = b + 999d
Hence, difference between their 1000th terms
= (a + 999d) (b + 999d) = a – b
= 100. [Using (i)]
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.
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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