-
Question 1: If the sum of first $m$ terms of an AP is the same as the sum of its first $n$ terms, show that the sum of its first $(m+n)$ terms is zero.
Solution:
Step 1: $S_m = S_n \Rightarrow \frac{m}{2}[2a + (m-1)d] = \frac{n}{2}[2a + (n-1)d]$.
Step 2: $2am + m(m-1)d = 2an + n(n-1)d \Rightarrow 2a(m-n) + d(m^2-m-n^2+n) = 0$.
Step 3: $2a(m-n) + d[(m-n)(m+n) - (m-n)] = 0$. Divide by $(m-n)$: $2a + d(m+n-1) = 0$.
Step 4: $S_{m+n} = \frac{m+n}{2}[2a + (m+n-1)d] = \frac{m+n}{2}(0) = 0$. -
Question 2: The ratio of the sums of first $m$ and first $n$ terms of an AP is $m^2 : n^2$. Show that the ratio of its $m^{th}$ and $n^{th}$ terms is $(2m-1) : (2n-1)$.
Solution:
Step 1: Let sum be $S_n = An^2 + Bn$. Since ratio is $m^2/n^2$, $B=0$, so $S_n = An^2$.
Step 2: $a_n = S_n - S_{n-1} = An^2 - A(n-1)^2 = A(n^2 - (n^2-2n+1)) = A(2n-1)$.
Step 3: Ratio $a_m : a_n = A(2m-1) : A(2n-1) = (2m-1) : (2n-1)$. -
Question 3: The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last terms to the product of the two middle terms is $7:15$. Find the numbers.
Solution:
Step 1: Let numbers be $a-3d, a-d, a+d, a+3d$. Sum $= 4a = 32 \Rightarrow a=8$.
Step 2: $\frac{(a-3d)(a+3d)}{(a-d)(a+d)} = \frac{7}{15} \Rightarrow \frac{a^2-9d^2}{a^2-d^2} = \frac{7}{15}$.
Step 3: $15(64-9d^2) = 7(64-d^2) \Rightarrow 960 - 135d^2 = 448 - 7d^2 \Rightarrow 128d^2 = 512 \Rightarrow d^2=4 \Rightarrow d=\pm 2$.
Answer: Numbers are 2, 6, 10, 14. -
Question 4: Solve the equation: $1 + 4 + 7 + 10 + \dots + x = 287$.
Solution:
Step 1: AP with $a=1, d=3$. Let $x$ be the $n^{th}$ term. $x = 1+(n-1)3 = 3n-2$.
Step 2: Sum $= \frac{n}{2}(1+x) = \frac{n}{2}(1+3n-2) = \frac{n(3n-1)}{2} = 287$.
Step 3: $3n^2 - n - 574 = 0$. Solving gives $n=14$.
Step 4: $x = 3(14) - 2 = 40$. -
Question 5: If the $p^{th}$ term of an AP is $1/q$ and the $q^{th}$ term is $1/p$, prove that the sum of first $pq$ terms is $\frac{1}{2}(pq+1)$.
Solution:
Step 1: $a + (p-1)d = 1/q$ and $a + (q-1)d = 1/p$. Subtracting: $(p-q)d = \frac{p-q}{pq} \Rightarrow d = \frac{1}{pq}$.
Step 2: Substitute $d$: $a = \frac{1}{pq}$.
Step 3: $S_{pq} = \frac{pq}{2}[2(\frac{1}{pq}) + (pq-1)\frac{1}{pq}] = \frac{pq}{2}[\frac{2+pq-1}{pq}] = \frac{1}{2}(pq+1)$. -
Question 6: Find the sum of all integers between 100 and 550 which are divisible by 9.
Solution:
Step 1: First term $a=108$, Last term $l=549$. $d=9$.
Step 2: $549 = 108 + (n-1)9 \Rightarrow 441 = 9(n-1) \Rightarrow n-1=49 \Rightarrow n=50$.
Step 3: $S_{50} = \frac{50}{2}(108+549) = 25(657) = 16425$. -
Question 7: If $S_n$ denotes the sum of first $n$ terms of an AP, prove that $S_{12} = 3(S_8 - S_4)$.
Solution:
Step 1: $S_8 = 4(2a+7d) = 8a+28d$. $S_4 = 2(2a+3d) = 4a+6d$.
Step 2: $S_8 - S_4 = 4a + 22d$.
Step 3: $3(S_8 - S_4) = 12a + 66d$.
Step 4: $S_{12} = 6(2a+11d) = 12a + 66d$. Hence proved. -
Question 8: The sum of the first $n$ terms of an AP is given by $S_n = 3n^2 - 4n$. Determine the AP and the $12^{th}$ term.
Solution:
Step 1: $a_1 = S_1 = 3-4 = -1$. $S_2 = 12-8 = 4$. $a_2 = S_2 - S_1 = 4 - (-1) = 5$.
Step 2: $d = 5 - (-1) = 6$. AP is $-1, 5, 11, \dots$
Step 3: $a_{12} = -1 + 11(6) = 65$. -
Question 9: If $\frac{a^n + b^n}{a^{n-1} + b^{n-1}}$ is the arithmetic mean between $a$ and $b$, then find the value of $n$.
Solution:
Step 1: $\frac{a^n + b^n}{a^{n-1} + b^{n-1}} = \frac{a+b}{2} \Rightarrow 2a^n + 2b^n = a^n + ab^{n-1} + ba^{n-1} + b^n$.
Step 2: $a^n - ba^{n-1} = ab^{n-1} - b^n \Rightarrow a^{n-1}(a-b) = b^{n-1}(a-b)$.
Step 3: $(a/b)^{n-1} = 1 = (a/b)^0 \Rightarrow n-1=0 \Rightarrow n=1$. -
Question 10: A thief runs with a uniform speed of 100 m/minute. After one minute, a policeman runs after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/minute every succeeding minute. After how many minutes the policeman will catch the thief?
Solution:
Step 1: Let policeman catch in $n$ mins. Distance $= \frac{n}{2}[200 + (n-1)10] = 5n^2 + 95n$.
Step 2: Thief runs for $n+1$ mins. Distance $= 100(n+1)$.
Step 3: $5n^2 + 95n = 100n + 100 \Rightarrow 5n^2 - 5n - 100 = 0 \Rightarrow n^2 - n - 20 = 0$.
Step 4: $(n-5)(n+4) = 0 \Rightarrow n=5$.