Practice Questions
Determine whether the sequence 5, 10, 20, 40… is AP, GP or HP.
Common ratio = 10/5 = 20/10 = 2. It is a GP.
If the 5th term of an AP is 18 and the 10th term is 33, find the common difference.
$a + 4d = 18$ and $a + 9d = 33$. Subtracting: $5d = 15 \Rightarrow d = 3$.
Insert 4 arithmetic means between 3 and 23. What is the common difference?
Total terms = $4 + 2 = 6$. $a = 3, a_6 = 23$. $23 = 3 + 5d \Rightarrow 20 = 5d \Rightarrow d = 4$.
If the 3rd term of a GP is 12 and the 6th term is 96, find the common ratio.
$ar^2 = 12$ and $ar^5 = 96$. $(ar^5)/(ar^2) = 96/12 \Rightarrow r^3 = 8 \Rightarrow r = 2$.
Find the sum of first 15 terms of the AP 7, 10, 13…
$a=7, d=3, n=15$. $S_{15} = 15/2 \times [14 + 14 \times 3] = 15/2 \times [14 + 42] = 15/2 \times 56 = 15 \times 28 = 420$.
If 1/a, 1/b, 1/c are in AP, then a, b, c are in:
By definition, if the reciprocals of terms form an AP, the terms themselves are in HP.
Find the 10th term of the GP 2, 6, 18, 54…
$a=2, r=3$. $T_{10} = 2 \times 3^{9} = 2 \times 19683 = 39366$.
If the sum of first n terms of an AP is $S_n = 3n^2 + 2n$, find the first term.
$S_1 = 3(1)^2 + 2(1) = 5$. First term = $S_1 = 5$.
Find the sum of an infinite GP: 1, 1/2, 1/4, 1/8…
$a=1, r=1/2$. $S = a/(1-r) = 1/(1-1/2) = 2$.
If three numbers in GP have product 27 and sum 13, find the middle term.
Let numbers be $a/r, a, ar$. Product = $a^3 = 27 \Rightarrow a = 3$.
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