Practice Questions
The sum of first n terms of an AP is $( 3n^2 + 5n )$. Find its nth term.
$a_1 = S_1 = 8$. $S_2 = 22 \Rightarrow a_2 = 14$. $d = 6$. $a_n = 8 + (n-1)6 = 6n + 2$.
Find the number of terms in the AP 5, 9, 13… that are less than 200.
$a+(n-1)d < 200 \Rightarrow 5+(n-1)4 < 200 \Rightarrow 4n+1 < 200 \Rightarrow 4n < 199 \Rightarrow n < 49.75$. Max $n = 49$.
Which term of the AP 4, 9, 14… is 124?
$124 = 4 + (n-1)5 \Rightarrow 120 = 5(n-1) \Rightarrow n-1 = 24 \Rightarrow n = 25$.
Find the sum of all two-digit numbers divisible by 7.
AP: 14, 21, ..., 98. $n=13$. Sum = $(13/2) \times (14+98) = 728$.
The 7th term of an AP is 34 and the 12th term is 54. Find the first term.
$a+6d = 34, a+11d = 54$. Subtracting: $5d=20 \Rightarrow d=4$. $a + 24 = 34 \Rightarrow a=10$.
How many terms of the AP 2, 5, 8… must be taken so that their sum is 406?
$406 = n/2 [4 + (n-1)3] \Rightarrow 812 = n(3n+1) \Rightarrow 3n^2 + n - 812 = 0$. Using quadratic formula or factors: $(n-14)(3n+58)=0 \Rightarrow n=14$.
Find three numbers in AP whose sum is 21 and product is 231.
Let numbers be $a-d, a, a+d$. Sum = $3a = 21 \Rightarrow a = 7$. Product = $(7-d) \times 7 \times (7+d) = 7(49-d^2) = 231 \Rightarrow 49-d^2 = 33 \Rightarrow d^2 = 16 \Rightarrow d = 4$. Numbers: 3, 7, 11.
The nth term of an AP is given by $a_n = 2n + 3$. Find the sum of first 20 terms.
$a_1 = 5$, $a_{20} = 43$. $S_{20} = 20/2 \times (5 + 43) = 10 \times 48 = 480$.
How many terms of the AP 20, 16, 12... must be taken so that the sum is maximum?
AP is decreasing. Terms become negative after $20 + (n-1)(-4) < 0 \Rightarrow n > 6$. So only first 5 terms are positive. Maximum sum with 5 terms.
The 4th term of an AP is 3 times the 1st term and the 7th term exceeds the 3rd term by 12. Find the common difference.
$a+3d = 3a \Rightarrow 3d = 2a$. $(a+6d)-(a+2d)=12 \Rightarrow 4d=12 \Rightarrow d=3$.
Premium Content
Unlock Arithmetic Progression Quiz and all premium lessons with a subscription.
From ₹199.99/year — See plans