Arithmetic Progressions — Important Questions

Answer: (b) $-4$.

Here $a_1 = 7 - 4 = 3$ and $a_2 = 7 - 8 = -1$, so $d = a_2 - a_1 = -1 - 3 = -4$. (The coefficient of $n$ always gives $d$.)

Answer: (b) $20$.

Here $a = -5$ and $d = \dfrac{5}{2}$. So $a_{11} = a + 10d = -5 + 10 \times \dfrac{5}{2} = -5 + 25 = 20$.

Answer: (b) $6$.

For an AP the middle term is the average of its neighbours: $2(x + 10) = 2x + (3x + 2) \Rightarrow 2x + 20 = 5x + 2 \Rightarrow 3x = 18 \Rightarrow x = 6$.

Answer: (b) $30$.

The numbers are $12, 15, 18, \dots, 99$, an AP with $a = 12$, $d = 3$ and $a_n = 99$. Then $99 = 12 + (n-1)3 \Rightarrow n - 1 = 29 \Rightarrow n = 30$.

Answer: (a) Both A and R are true and R is the correct explanation of A.

Using $S_n = \dfrac{n(n+1)}{2}$ with $n = 20$ gives $S_{20} = \dfrac{20 \times 21}{2} = 210$, so R directly proves A.

Here $a = 27$, $d = -3$. Using $S_n = \dfrac{n}{2}[2a + (n-1)d] = 0$:

$\dfrac{n}{2}[54 + (n-1)(-3)] = 0 \Rightarrow 54 - 3(n-1) = 0 \Rightarrow 57 = 3n \Rightarrow n = 19$.

So 19 terms must be taken.

Here $a = 3$, $d = 5$ and $a_n = 78$.

$78 = 3 + (n-1)5 \Rightarrow 75 = 5(n-1) \Rightarrow n - 1 = 15 \Rightarrow n = 16$.

Hence $78$ is the 16th term.

$S_7 = \dfrac{7}{2}[2a + 6d] = 49 \Rightarrow a + 3d = 7$ ...(i)

$S_{17} = \dfrac{17}{2}[2a + 16d] = 289 \Rightarrow a + 8d = 17$ ...(ii)

Subtracting (i) from (ii): $5d = 10 \Rightarrow d = 2$, so $a = 7 - 6 = 1$.

$S_n = \dfrac{n}{2}[2(1) + (n-1)2] = \dfrac{n}{2}(2n) = n^2$.

Therefore $S_{20} = 20^2 = \mathbf{400}$.

$a_4 + a_8 = (a + 3d) + (a + 7d) = 2a + 10d = 24 \Rightarrow a + 5d = 12$ ...(i)

$a_6 + a_{10} = (a + 5d) + (a + 9d) = 2a + 14d = 44 \Rightarrow a + 7d = 22$ ...(ii)

Subtracting (i) from (ii): $2d = 10 \Rightarrow d = 5$, so $a = 12 - 25 = -13$.

The first three terms are $-13,\ -8,\ -3$.

The left side is an AP with $a = 1$, $d = 3$ and last term $x$, whose sum is $287$.

$S_n = \dfrac{n}{2}[2(1) + (n-1)3] = 287 \Rightarrow n(3n - 1) = 574 \Rightarrow 3n^2 - n - 574 = 0$.

$n = \dfrac{1 + \sqrt{1 + 6888}}{6} = \dfrac{1 + 83}{6} = 14$ (taking the positive root).

Then $x = a_{14} = 1 + 13 \times 3 = \mathbf{40}$.

Here $a = 20$, $d = -1$ and $S_n = 200$.

$\dfrac{n}{2}[2(20) + (n-1)(-1)] = 200 \Rightarrow n(41 - n) = 400 \Rightarrow n^2 - 41n + 400 = 0$.

$(n - 16)(n - 25) = 0 \Rightarrow n = 16$ or $n = 25$.

If $n = 25$, then $a_{25} = 20 + 24(-1) = -4$, which is impossible (a row cannot hold a negative number of logs). So $n = 16$.

Top row logs $= a_{16} = 20 + 15(-1) = 5$.

Hence the logs are placed in 16 rows with 5 logs in the top row.

Given $\dfrac{S_m}{S_n} = \dfrac{m^2}{n^2}$:

$\dfrac{\frac{m}{2}[2a + (m-1)d]}{\frac{n}{2}[2a + (n-1)d]} = \dfrac{m^2}{n^2} \Rightarrow \dfrac{2a + (m-1)d}{2a + (n-1)d} = \dfrac{m}{n}$.

Cross-multiplying: $n[2a + (m-1)d] = m[2a + (n-1)d]$

$\Rightarrow 2an + mnd - nd = 2am + mnd - md \Rightarrow 2a(n - m) = d(n - m)$.

Since $m \neq n$, we get $d = 2a$.

Now $\dfrac{a_m}{a_n} = \dfrac{a + (m-1)d}{a + (n-1)d} = \dfrac{a + (m-1)(2a)}{a + (n-1)(2a)} = \dfrac{a(2m - 1)}{a(2n - 1)} = \dfrac{2m - 1}{2n - 1}$.

Hence the ratio of the $m$th to the $n$th term is $(2m - 1) : (2n - 1)$.

Let first-year production $= a$ and annual increase $= d$.

Given $a_3 = a + 2d = 6000$ and $a_7 = a + 6d = 7000$.

Subtracting: $4d = 1000 \Rightarrow d = 250$.

(i) $a = 6000 - 2(250) = 6000 - 500 = \mathbf{5500}$ sets.

(ii) Annual increase $d = \mathbf{250}$ sets per year.

(iii) Production in the $10$th year $= a_{10} = 5500 + 9(250) = 5500 + 2250 = \mathbf{7750}$ sets.

OR Total in first $7$ years $= S_7 = \dfrac{7}{2}[2(5500) + 6(250)] = \dfrac{7}{2}[11000 + 1500] = \dfrac{7}{2}(12500) = \mathbf{43750}$ sets.