Arithmetic Progressions — CBSE Class 10 Maths Important Questions

Answer: (b) $-4$-4.

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

Answer: (b) $20$20.

Here $a = -5$a = -5 and $d = \dfrac{5}{2}$d = 5/2. So $a_{11} = a + 10d = -5 + 10 \times \dfrac{5}{2} = -5 + 25 = 20$a_11 = a + 10d = -5 + 10 × 5/2 = -5 + 25 = 20.

Answer: (b) $6$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$2(x + 10) = 2x + (3x + 2) 2x + 20 = 5x + 2 3x = 18 x = 6.

Answer: (b) $30$30.

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

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

Using $S_n = \dfrac{n(n+1)}{2}$S_n = n(n+1)/2 with $n = 20$n = 20 gives $S_{20} = \dfrac{20 \times 21}{2} = 210$S_20 = 20 × 21/2 = 210, so R directly proves A.

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

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

So 19 terms must be taken.

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

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

Hence $78$78 is the 16th term.

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

$S_{17} = \dfrac{17}{2}[2a + 16d] = 289 \Rightarrow a + 8d = 17$S_17 = 17/2[2a + 16d] = 289 a + 8d = 17 ...(ii)

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

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

Therefore $S_{20} = 20^2 = \mathbf{400}$S_20 = 20^2 = 400.

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

$a_6 + a_{10} = (a + 5d) + (a + 9d) = 2a + 14d = 44 \Rightarrow a + 7d = 22$a_6 + a_10 = (a + 5d) + (a + 9d) = 2a + 14d = 44 a + 7d = 22 ...(ii)

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

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

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

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

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

Then $x = a_{14} = 1 + 13 \times 3 = \mathbf{40}$x = a_14 = 1 + 13 × 3 = 40.

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

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

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

If $n = 25$n = 25, then $a_{25} = 20 + 24(-1) = -4$a_25 = 20 + 24(-1) = -4, which is impossible (a row cannot hold a negative number of logs). So $n = 16$n = 16.

Top row logs $= a_{16} = 20 + 15(-1) = 5$= a_16 = 20 + 15(-1) = 5.

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

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

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

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

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

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

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

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

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

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

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

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

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

(iii) Production in the $10$10th year $= a_{10} = 5500 + 9(250) = 5500 + 2250 = \mathbf{7750}$= a_10 = 5500 + 9(250) = 5500 + 2250 = 7750 sets.

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