JEE Mathematics Sequence and Series MCQs Set 45

Answer: (d) Both are true and Reason explains Assertion

Explanation:
1. \( a + b = 68 \) and ab = 256, so a and \( b \) are roots of \( t^2 - 68t + 256 = 0 \), which gives 4 and 64.
2. This uses the reason's formulas.

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (d) Both are true and Reason explains Assertion

Explanation:
1. \( S(n) \) proportional to \( n^2 \) gives \( t(n) \) proportional to \( 2n - 1 \), so \( T5/T2 = 9/3 = 3 \).
2. The nth-term formula is the basis for \( t(n) = S(n) - S(n - 1) \).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (c) \(a=3,\ r=\dfrac14\)

Explanation:
1. \(\dfrac a{1-r}=4\) and \(ar=\dfrac34\), so \(4r(1-r)=\dfrac34\), i.e. \(16r^2-16r+3=0\).
2. \(r=\dfrac14\) or \(\dfrac34\) with \(a=3\) or \(1\); of the options only \(a=3,r=\frac14\) fits.

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (a) 384

Explanation:
1. \(3072=3\cdot2^{10}\), so there are 11 terms.
2. The 4th from the end is the 8th term \(3\cdot2^7=384\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (c) A.P.

Explanation:
1. \(x\log a=y\log b=z\log c=k\), so \(\dfrac1x=\dfrac{\log a}k\) and so on.
2. \(\log a,\log b,\log c\) are in \( A.P \). (since a, \( b, c \) are in \( G.P. \)), hence so are \(\dfrac1x,\dfrac1y,\dfrac1z\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (c) \(a\ge12\)

Explanation:
1. \(a=5^{1+x}+5^{1-x}+5^{2x}+5^{-2x}\).
2. By \( A.M.-G.M \). each pair is at least \(2\sqrt{5\cdot5}=10\) and \(2\), so \(a\ge12\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (a) never less than n

Explanation:
1. By \( A.M.-G.M \)., (sum)/n >= \( (product)^{1/n} = 1 \), so the sum is at least \( n \).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (c) \(n+2^{-n}-1\)

Explanation:
1. The k-th term is \(1-2^{-k}\).
2. Sum \(=n-\left(1-2^{-n}\right)=n+2^{-n}-1\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (d) 330

Explanation:
1. The middle term is the 6th, the average of the 11 terms, so the sum is \( 11 x 30 = 330 \).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (c) \(-4,\ 2\)

Explanation:
1. Let the numbers be \( x, y, z, w \) with \( w = x \). Then \( z = y + 6 \) and \( w = y + 12 \), so \( x = y + 12 \).
2. \( G.P \).: \( y^2 \) = xz = \( (y + 12)(y + 6) \) gives \( 18y + 72 = 0 \), so \( y = -4, z = 2 \) (and \( x = w = 8 \)).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (c) \(\text{the series may or may not be arithmetic}\)

Explanation:
1. \(a_n=S_n-S_{n-1}=2an-a+b\) for \(n\ge2\), an \( A.P \). from the 2nd term onwards.
2. \(a_1=a+b+c\) fits the same formula only when \(c=0\); so the series is an \( A.P \). exactly when \(c=0\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (a) 1 : 2

Explanation:
1. \(5(2a+9d)=4\cdot\dfrac52(2a+4d)\) gives \(10a+45d=20a+40d\), so \(5d=10a\).
2. \(a:d=1:2\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (b) \(2xyz\)

Explanation:
1. \(g_1=y^{2/3}z^{1/3}\) and \(g_2=y^{1/3}z^{2/3}\), so \(g_1^3+g_2^3=y^2z+yz^2=yz(y+z)\).
2. \(y+z=2x\), so the sum is \(2xyz\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (c) 1,71,650

Explanation:
1. \(\sum_{k=1}^{50}(4k^2-1)=4\cdot\dfrac{50\cdot51\cdot101}6-50=171700-50=171650\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (a) 191

Explanation:
1. The common terms are 11, 31, 51, ... with common difference lcm(4, 5) = 20.
2. 10th common term = \( 11 + 9 x 20 = 191 \).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (c) 20 years

Explanation:
1. \(\dfrac n2[270+15(n-1)]=5550\) gives \(n^2+17n-740=0\).
2. \((n-20)(n+37)=0\), so \(n=20\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (b) G.P.

Explanation:
1. The differences of the given terms are \(\dfrac{b-a}{D}\) and \(\dfrac{c-b}{D}\) with the same \(D=(\sqrt a+\sqrt b)(\sqrt b+\sqrt c)(\sqrt c+\sqrt a)\); equal differences need \(b-a=c-b\), so \(a,b,c\) are in \( A.P \).
2. Then \(ax+1,bx+1,cx+1\) are in \( A.P \). and the powers of 9 are in \( G.P \). for every \(x\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (a) \((2+\sqrt3):(2-\sqrt3)\)

Explanation:
1. \(\dfrac{a+b}2=2\sqrt{ab}\); with \(t=\sqrt{a/b}\): \(t^2-4t+1=0\), \(t=2\pm\sqrt3\).
2. \(\dfrac ab=t^2=(2+\sqrt3)^2=\dfrac{2+\sqrt3}{2-\sqrt3}\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (c) 0

Explanation:
1. \(\dfrac{S_i}{n_i}=a+\dfrac{(n_i-1)d}2\) is a linear function \(p+qn_i\).
2. The expression is \(\sum(p+qn_i)(n_j-n_k)\) over cyclic indices, and both \(\sum(n_j-n_k)\) and \(\sum n_i(n_j-n_k)\) are zero.

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (c) 446

Explanation:
1. The k-th term is \(\dfrac{(k(k+1)/2)^2}{k^2}=\dfrac{(k+1)^2}4\).
2. \(\sum_{k=1}^{16}\dfrac{(k+1)^2}4=\dfrac{1785-1}4=446\) (the sum of squares up to 17 is 1785).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Answer: (a) \(\dfrac{n^2(n+1)}2\)

Explanation:
1. For odd \(n\), the sum is the even-case sum for \(n-1\) plus the last term \(n^2\).
2. \(\dfrac{(n-1)n^2}2+n^2=\dfrac{n^2(n+1)}2\) (check: \( n = 1 \) gives 1 and \( n = 3 \) gives 18).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.