Progressions (A.P., G.P., H.P.) Questions and Answers

Answer: C

\[ S_{3} = a(1 + r + r^{2}) = \frac{35}{4} \]

\[ 1 + r + r^{2} = \frac{7}{4} \quad\Longrightarrow\quad 4r^{2} + 4r - 3 = 0 \]

\[ r = \frac{-4 \pm \sqrt{16 + 48}}{8} = \frac{-4 \pm 8}{8} \]

\[ r = \frac{1}{2} \quad\text{or}\quad r = -\frac{3}{2} \]

Both roots are valid common ratios, so both answers apply.

Answer: A

The series is 100, 105, ..., 200. This is an A.P. with \(a=100, l=200, d=5\). Number of terms \(n = \frac{l-a}{d} + 1 = \frac{200-100}{5} + 1 = 20+1=21\). Sum = \(\frac{n}{2}(a+l) = \frac{21}{2}(100+200) = \frac{21}{2}(300) = 21 \times 150 = 3150\).

Answer: B

Given \(a+8d=499\) and \(a+498d=9\). Subtracting gives \(-490d = 490 \Rightarrow d=-1\). Substituting back, \(a+8(-1)=499 \Rightarrow a=507\). We want to find n for which \(a_n=0\). \(a_n = a+(n-1)d = 0 \Rightarrow 507 + (n-1)(-1) = 0 \Rightarrow 507 = n-1 \Rightarrow n=508\). The 508th term is zero.

Answer: A

We have \(a_5 = ar^4 = 81\) and \(a_2 = ar = 24\). Dividing the equations: \(\frac{ar^4}{ar} = \frac{81}{24} \Rightarrow r^3 = 27/8\). So, \(r=3/2\). Substitute r back into the second equation: \(a(3/2) = 24 \Rightarrow a = 24 \times 2/3 = 16\). The G.P. is 16, \(16 \times 3/2 = 24\), \(24 \times 3/2 = 36\), ...

Answer: B

Given \(S_p = S_q\). \(\frac{p}{2}[2a+(p-1)d] = \frac{q}{2}[2a+(q-1)d]\). After simplification, we get \(2a(p-q) + d(p^2-p-q^2+q)=0 \Rightarrow (p-q)[2a+d(p+q-1)]=0\). Since p is not equal to q, \(2a+d(p+q-1)=0\). The sum of (p+q) terms is \(S_{p+q} = \frac{p+q}{2}[2a+(p+q-1)d]\). Since the term in the bracket is 0, the sum is 0.

Answer: B

The ancestors form a geometric progression with \( a = 2 \), \( r = 2 \).

\[ S_{10} = \frac{2(2^{10} - 1)}{2 - 1} = 2(1024 - 1) \]

\[ = 2046 \]

2048 is \( 2^{11} \), two more than the sum - the nearest trap.

Answer: B

We have \(\frac{S_3}{S_6} = \frac{125}{152}\). Let the sum formula be \(S_n=a(r^n-1)/(r-1)\). So, \(\frac{a(r^3-1)/(r-1)}{a(r^6-1)/(r-1)} = \frac{r^3-1}{r^6-1} = \frac{r^3-1}{(r^3-1)(r^3+1)} = \frac{1}{r^3+1}\). We have \(\frac{1}{r^3+1} = \frac{125}{152} \Rightarrow r^3+1 = \frac{152}{125} \Rightarrow r^3 = \frac{152}{125} - 1 = \frac{27}{125}\). So \(r = 3/5\).

Answer: C

Correct answer: 127

First count the terms. With \(a = 7\), \(d = 6\) and last term \(l = 247\):

\[247 = 7 + (n-1)6 \quad\Longrightarrow\quad 240 = 6(n-1) \quad\Longrightarrow\quad n = 41\]

With an odd number of terms there is a single middle term, the \(\tfrac{41+1}{2} = 21\)st.

\[a_{21} = 7 + 20 \times 6 = 127\]

As a check, 127 sits exactly halfway between the ends: \(\tfrac{7 + 247}{2} = 127\).

Answer: B

The sum of the squares of the first n natural numbers is given by the formula \(S_n = \frac{n(n+1)(2n+1)}{6}\). Here, n=10. \(S_{10} = \frac{10(10+1)(2 \times 10+1)}{6} = \frac{10 \times 11 \times 21}{6} = 5 \times 11 \times 7 = 385\).

Answer: A

The corresponding A.P. has a 3rd term of 7 and a 5th term of 11. So, \(a+2d=7\) and \(a+4d=11\). Subtracting the equations gives \(2d=4\), so \(d=2\). Substituting back, \(a+2(2)=7 \Rightarrow a+4=7 \Rightarrow a=3\). The first term of the A.P. is 3. The first term of the H.P. is the reciprocal, 1/3.