IPMPrepThe 3rd term of a GP is 9 and its 6th term is 243. Find the sum of its first 7 terms, given that the common ratio is positive.
Solution
r3 = 2439 = 27, so r = 3; first term = 99 = 1.
S7 = 1((3)7 − 1)/(3 − 1) = 1093.
Answer: 1093
Quantitative Ability · Algebra
Sequences and Series: IPMAT practice questions
24 questions, 7 of them short answer (type the number, as in IPMAT Indore). Each is marked with the exam level it suits. Answer, then open the step-by-step solution.
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Find the sum to infinity: 1(15) + 2(15)2 + 3(15)3 + …
Solution
For |x| < 1, Σ k xk = x/(1 − x)2.
= 15 / (45)2 = 516.
Answer: (D) 516
Find the sum to infinity: 1(14) + 2(14)2 + 3(14)3 + …
Solution
For |x| < 1, Σ k xk = x/(1 − x)2.
= 14 / (34)2 = 49.
Answer: (D) 49
Find 1/(1 × 2) + 1/(2 × 3) + 1/(3 × 4) + … + 1/(10 × 11).
Solution
1/(k(k + 1)) = 1/k − 1/(k + 1), so the sum telescopes.
Sum = 1 − 111 = 1011.
Answer: (C) 1011
An infinite GP has first term 8 and sum 323. Find its common ratio.
Solution
S = a/(1 − r): 323 = 8/(1 − r).
1 − r = 8 ÷ 323 = 34, so r = 14.
Answer: (A) 14
The 3rd term of a GP is 16 and its 6th term is 128. Find the sum of its first 9 terms, given that the common ratio is positive.
Solution
r3 = 12816 = 8, so r = 2; first term = 164 = 4.
S9 = 4((2)9 − 1)/(2 − 1) = 2044.
Answer: (D) 2044
Find 12 + 22 + 32 + … + 172.
Solution
Σk2 = n(n + 1)(2n + 1)/6 = 17 × 18 × 356 = 1785.
Answer: (C) 1785
The 7th term of an AP is 20 and its 6th term is 17. Find the sum of its first 28 terms.
Solution
d = (20 − 17)/(7 − 6) = 3; first term a = 20 − 6 × 3 = 2.
S28 = (282)[2 × 2 + 27 × 3] = 1190.
Answer: 1190
Find 1/(1 × 2) + 1/(2 × 3) + 1/(3 × 4) + … + 1/(11 × 12).
Solution
1/(k(k + 1)) = 1/k − 1/(k + 1), so the sum telescopes.
Sum = 1 − 112 = 1112.
Answer: (D) 1112
Find 13 + 23 + 33 + … + 223.
Solution
Σk3 = [n(n + 1)/2]2 = 2532 = 64009.
Answer: 64009
How many integers from 200 to 800 are divisible by 13?
Solution
The first is 208 and the last is 793; they form an AP with common difference 13.
Number of terms = (793 − 208)/13 + 1 = 46.
Answer: (B) 46
What is the sum of all integers from 100 to 500 that are divisible by 8?
Solution
The first is 104 and the last is 496; they form an AP with common difference 8.
Number of terms = (496 − 104)/8 + 1 = 50.
Sum = 50 × (104 + 496)/2 = 15000.
Answer: (A) 15,000
The arithmetic mean of two positive numbers is 13 and their geometric mean is 12. Find the larger number.
Solution
a + b = 26 and ab = 144.
They are the roots of t2 − 26t + 144 = 0: 8 and 18.
The larger is 18.
Answer: 18
Find 1 × 2 + 2 × 3 + 3 × 4 + … + 13 × 14.
Solution
k(k + 1) = k2 + k, and the sum works out to n(n + 1)(n + 2)/3.
= 13 × 14 × 153 = 910.
Answer: (C) 910
The 8th term of an AP is −2 and its 10th term is −4. Find the sum of its first 18 terms.
Solution
d = (−2 − (−4))/(8 − 10) = −1; first term a = −2 − 7 × (−1) = 5.
S18 = (182)[2 × 5 + 17 × (−1)] = −63.
Answer: (A) −63
Find the harmonic mean of 4 and 16.
Solution
HM = 2ab/(a + b) = 2 × 4 × 1620 = 325.
Check: GM2 = AM × HM, and 82 = 10 × 325.
Answer: (D) 6.4
How many integers from 200 to 800 are divisible by 7?
Solution
The first is 203 and the last is 798; they form an AP with common difference 7.
Number of terms = (798 − 203)/7 + 1 = 86.
Answer: 86
Find the sum to infinity: 1(23) + 2(23)2 + 3(23)3 + …
Solution
For |x| < 1, Σ k xk = x/(1 − x)2.
= 23 / (13)2 = 6.
Answer: (B) 6
The arithmetic mean of two positive numbers is 132 and their geometric mean is 6. Find the larger number.
Solution
a + b = 13 and ab = 36.
They are the roots of t2 − 13t + 36 = 0: 4 and 9.
The larger is 9.
Answer: 9
An infinite GP has first term 4 and sum 5. Find its common ratio.
Solution
S = a/(1 − r): 5 = 4/(1 − r).
1 − r = 4 ÷ 5 = 45, so r = 15.
Answer: (A) 15
The 12th term of an AP is −11 and its 8th term is −7. Find the sum of its first 21 terms.
Solution
d = (−11 − (−7))/(12 − 8) = −1; first term a = −11 − 11 × (−1) = 0.
S21 = (212)[2 × 0 + 20 × (−1)] = −210.
Answer: −210
Find the sum of the infinite GP 18, 185, 1825, …
Solution
First term 18, common ratio 15.
Sum = 18/(1 − (15)) = 452.
Answer: (A) 452
The 3rd term of a GP is 20 and its 7th term is 320. Find the sum of its first 7 terms, given that the common ratio is negative.
Solution
r4 = 32020 = 16, so r = −2; first term = 204 = 5.
S7 = 5((−2)7 − 1)/(−2 − 1) = 215.
Answer: (C) 215
Find 1/(1 × 3) + 1/(3 × 5) + 1/(5 × 7) + … + 1/(13 × 15).
Solution
1/((2k − 1)(2k + 1)) = (12)[1/(2k − 1) − 1/(2k + 1)].
Sum = (12)(1 − 115) = 715.
Answer: (C) 715
These are original practice questions written by Target Campus in the pattern, syllabus and difficulty of IPMAT Indore, IPMAT Rohtak, JIPMAT, NPAT and SET. They are not previous-year papers.