Algebra – Sequence & Series (AP, GP, Binomial Theorem) Quiz - ICSE Class 11 Mathematics
Free practice quiz: Algebra – Sequence & Series (AP, GP, Binomial Theorem) (ICSE Class 11 Mathematics). 10 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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The sum of the first n terms of an arithmetic progression is given by $S_n = 2n^2 + 3n$. If the common difference of the AP is $d$, find the value of $d$.
- 2
- 3
- 4
- 5
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Answer: C. 4
We know that $a_n = S_n - S_{n-1}$. For $n=1$, $a_1 = S_1 = 2(1)^2 + 3(1) = 5$. For $n=2$, $a_1 + a_2 = S_2 = 2(2)^2 + 3(2) = 8 + 6 = 14$. Thus, $a_2 = S_2 - a_1 = 14 - 5 = 9$. The common difference $d = a_2 - a_1 = 9 - 5 = 4$. Alternatively, in an AP, $S_n = \frac{n}{2}[2a + (n-1)d] = \frac{d}{2}n^2 + (a - \frac{d}{2})n$. Comparing this with $S_n = 2n^2 + 3n$, we get $\frac{d}{2} = 2$, which implies $d = 4$.
The sum of $n$ terms of an arithmetic progression is $S_n = 3n^2 - n$. Find the common difference of the AP.
- 2
- 3
- 4
- 6
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Answer: D. 6
The first term $a_1 = S_1 = 3(1)^2 - 1 = 2$. The sum of the first two terms $S_2 = 3(2)^2 - 2 = 12 - 2 = 10$. The second term $a_2 = S_2 - S_1 = 10 - 2 = 8$. The common difference $d = a_2 - a_1 = 8 - 2 = 6$. Alternatively, for an AP, $S_n = \frac{d}{2}n^2 + (a - \frac{d}{2})n$. Comparing $S_n = 3n^2 - n$ with the general form, $\frac{d}{2} = 3$, which implies $d = 6$.
Find the coefficient of $x^3$ in the expansion of $(1 + 2x)^7$.
- 140
- 280
- 320
- 420
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Answer: B. 280
The general term in the binomial expansion of $(a+b)^n$ is $T_{r+1} = inom{n}{r} a^{n-r} b^r$. Here, $a=1$, $b=2x$, and $n=7$. We want the coefficient of $x^3$, so the power of $b$ should be 3, which means $r=3$. The term is $T_{3+1} = T_4 = inom{7}{3} (1)^{7-3} (2x)^3$. The coefficient of $x^3$ is $inom{7}{3} (1)^4 (2)^3$. $inom{7}{3} = \frac{7!}{3!4!} = \frac{7 imes 6 imes 5}{3 imes 2 imes 1} = 35$. $(2)^3 = 8$. So the coefficient is $35 imes 8 = 280$.
The sum of the first 3 terms of a GP is 13, and the sum of the squares of these terms is 91. Find the common ratio.
- 2
- 3
- -2
- -3
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Answer: B. 3
Let the first three terms of the GP be $a/r, a, ar$. Their sum is $a/r + a + ar = a(1/r + 1 + r) = 13$. The sum of their squares is $(a/r)^2 + a^2 + (ar)^2 = a^2(1/r^2 + 1 + r^2) = 91$. From the first equation, $a(\frac{1+r+r^2}{r}) = 13$. From the second equation, $a^2(\frac{1+r^2+r^4}{r^2}) = 91$. We can also write $1+r^2+r^4 = (1+r^2)^2 - r^2 = (1+r^2-r)(1+r^2+r)$. So, $a^2 \frac{(1+r^2-r)(1+r^2+r)}{r^2} = 91$. From $a(\frac{1+r+r^2}{r}) = 13$, we have $a^2 (\frac{1+r+r^2}{r})^2 = 169$. $a^2 \frac{(1+r+r^2)^2}{r^2} = 169$. Dividing the second sum equation by the square of the first sum equation (after some manipulation is required). A simpler approach: Let the terms be $a, ar, ar^2$. Then $a+ar+ar^2 = 13$ and $a^2 + a^2r^2 + a^2r^4 = 91$. $a(1+r+r^2) = 13$ and $a^2(1+r^2+r^4) = 91$. $a^2(1+r^2+r^4) = a^2(1+r+r^2)(1-r+r^2) = 91$. Substitute $a(1+r+r^2)=13$. So, $13a(1-r+r^2) = 91$, which means $a(1-r+r^2) = 7$. We have two equations: 1) $a(1+r+r^2) = 13$ and 2) $a(1-r+r^2) = 7$. Divide equation 1 by equation 2: $\frac{1+r+r^2}{1-r+r^2} = \frac{13}{7}$. $7(1+r+r^2) = 13(1-r+r^2)$. $7+7r+7r^2 = 13-13r+13r^2$. $6r^2 - 20r + 6 = 0$. $3r^2 - 10r + 3 = 0$. Factoring, $(3r-1)(r-3) = 0$. So $r=3$ or $r=1/3$. If $r=3$, then $a(1+3+9) = 13$, $13a=13$, so $a=1$. The terms are $1/3, 1, 3$. Sum of squares: $(1/3)^2 + 1^2 + 3^2 = 1/9 + 1 + 9 = 10 + 1/9 = 91/9 eq 91$. Error in logic. Let's use the terms $a, ar, ar^2$. Sum: $a(1+r+r^2)=13$. Sum of squares: $a^2(1+r^2+r^4)=91$. Let's try the earlier representation: $a/r, a, ar$. Sum: $a(1/r+1+r)=13$. Sum of squares: $a^2(1/r^2+1+r^2)=91$. Divide the second equation by the first squared: $\frac{a^2(1/r^2+1+r^2)}{a^2(1/r+1+r)^2} = \frac{91}{169}$. $\frac{(1/r^2+1+r^2)}{(1/r+1+r)^2} = \frac{7}{13}$. $\frac{\frac{1+r^2+r^4}{r^2}}{\frac{(1+r+r^2)^2}{r^2}} = \frac{7}{13}$. $\frac{1+r^2+r^4}{(1+r+r^2)^2} = \frac{7}{13}$. $\frac{(1+r^2-r)(1+r^2+r)}{(1+r+r^2)^2} = \frac{7}{13}$. $\frac{1-r+r^2}{1+r+r^2} = \frac{7}{13}$. $13(1-r+r^2) = 7(1+r+r^2)$. $13-13r+13r^2 = 7+7r+7r^2$. $6r^2 - 20r + 6 = 0$. $3r^2 - 10r + 3 = 0$. $(3r-1)(r-3)=0$. So $r=3$ or $r=1/3$. If $r=3$, $a(1/3+1+3)=13 \implies a(13/3)=13 \implies a=3$. Terms are $1, 3, 9$. Sum = $1+3+9=13$. Sum of squares = $1^2+3^2+9^2 = 1+9+81=91$. This matches. So $r=3$. If $r=1/3$, $a(3+1+1/3)=13 \implies a(13/3)=13 \implies a=3$. Terms are $9, 3, 1$. Sum = $9+3+1=13$. Sum of squares = $9^2+3^2+1^2 = 81+9+1=91$. This also matches. Both $r=3$ and $r=1/3$ are valid. The options provided are 2, 3, -2, -3. Only 3 is present. Thus, the common ratio is 3.
Find the sum of the series $1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \dots + n(n+1)$
- n(n+1)(n+2)/3
- n(n+1)(n+2)/2
- n(n+1)(n+2)/4
- n(n+1)(n+2)/5
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Answer: A. n(n+1)(n+2)/3
The general term of the series is $T_k = k(k+1) = k^2 + k$. The sum of the series is $S_n = \sum_{k=1}^n T_k = \sum_{k=1}^n (k^2 + k) = \sum_{k=1}^n k^2 + \sum_{k=1}^n k$. We know the formulas for the sum of the first $n$ integers and the sum of the first $n$ squares: $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ and $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$. Substituting these into the sum: $S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$. To simplify, find a common denominator: $S_n = \frac{n(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6}$. Factor out $\frac{n(n+1)}{6}$: $S_n = \frac{n(n+1)}{6} [(2n+1) + 3]$. $S_n = \frac{n(n+1)}{6} (2n+4)$. $S_n = \frac{n(n+1)2(n+2)}{6}$. $S_n = \frac{n(n+1)(n+2)}{3}$.