Question types

Sequences and Series question types

58 questions across 7 question groups — pick any mix to generate a Applied Maths paper with step-by-step answer keys.

58
Questions
7
Question groups
5
Question types
Sample Questions

Sequences and Series questions

One sample from each question group in this chapter. Select any group above to see the full set with answer keys.

Q 1MCQ1 Mark
The minimum value of the expression $3^x+3^{1-x}$, $x \in R$, is
  • A
    $0$
  • B
    $\frac{1}{3}$
  • C
    3
  • $2 \sqrt{3}$

Answer: D.

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Q 2MCQ1 Mark
Let $S$ be the sum, $P$ be the product and $R$ be the sum of the reciprocals of 3 terms of a G.P. then $P^2 R^3$ : $S^3$ in equal to
  • $1: 1$
  • B
    (common ratio) ${ }^n: 1$
  • C
    $(\text { (first term })^2:(\text { common ratio })^2$
  • D
    none of these

Answer: A.

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Q 3MCQ1 Mark
In a G.P. of even number of terms, the sum of all terms is 5 times the sum of the odd terms. The common ratio of the G.P. is
  • A
    $\frac{-4}{5}$
  • B
    $\frac{1}{5}$
  • 4
  • D
    none of these

Answer: C.

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Q 4MCQ1 Mark
In a G.P. of positive terms, if any term in equal to the sum of the next two terms. Then the common ratio of the G.P. is
  • $\frac{(\sqrt{5}-1)}{2}$
  • B
    $\frac{\sqrt{5}+1}{2}$
  • C
    $\frac{-\sqrt{5}-1}{2}$
  • D
    $\frac{ \pm \sqrt{5}-1}{2}$

Answer: A.

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Q 5MCQ1 Mark
If the sum of $n$ terms of an A.P. is given by $S_n=3 n$ $+2 n^2$, then the common difference of the A.P. is
  • A
    3
  • B
    2
  • C
    6
  • 4

Answer: D.

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Q 153 Marks Question3 Marks
The first term of an A.P. is $a$ and the sum of the first $p$ terms is zero, show that the sum of its next $q$ terms is $\frac{-a(p+q) q}{p-1}$.
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If $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of an A.P. and G.P. are both $a, b$ and $c$ respectively, then show that $a^{b-c}$. $b^{c-a} \cdot c^{a-b}=1$.
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If $p, q, r$ are in G.P. and the equations $p x^2+2 q x+$ $r=0$ and $d x^2+2 e x+f=0$ have a common root, then show that $\frac{d}{p}, \frac{e}{q}, \frac{f}{r}$ are in A.P.
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The product of first three terms of G.P. is 1000 . If 6 is added to its second term and 7 is added to its third term, the terms become in A.P. Find G.P.
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If $a, b, c$ are in A.P. show that $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right)$, $c\left(\frac{1}{a}+\frac{1}{b}\right)$ are also in A.P.
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Column - IColumn - II
(a) $1^2+2^2+3^2+\ldots+n^2$(i) $\left(\frac{n(n+1)}{2}\right)^2$
(b) $1^3+2^3+3^3+\ldots+n^3$(ii) n(n+1)
(c) $2+4+6+\ldots+2 n$(iii) $\frac{n(n+1)(2 n+1)}{6}$
(d) $1+2+3+\ldots+n$(iv) $\frac{n(n+1)}{2}$
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Column - IColumn - II
(a) 2, 3, 5, 7, . . .(i) series
(b) 13, 8, 3, -2, -7, . . .(ii) sequence
(c) 1 + 2 + 3 + 4 + . . .(iii) A.P.
(d) 1, 3, 5, 7, . . .(iv) Sequence with $a_n=2 n-1$
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