Arithmetic Progressions — NCERT Solutions, AI Tutor & Practice

To be an AP, the difference between consecutive terms must be constant.

• $a_2 - a_1 = 7 - 3 = 4$
• $a_3 - a_2 = 11 - 7 = 4$
• $a_4 - a_3 = 15 - 11 = 4$

Since the difference is the same each time, the list is an AP with common difference $d = 4$.

Fare for 1 km $= 15$.

Each additional km adds ₹8:

• 2 km: $15 + 8 = 23$
• 3 km: $23 + 8 = 31$
• 4 km: $31 + 8 = 39$

Sequence: $15, 23, 31, 39, \dots$

Differences: $23-15 = 8$, $31-23 = 8$, $39-31 = 8$.

Since the common difference is constant, the fares form an AP with $a = 15$ and $d = 8$.

The first term is the very first number: $a = -5$

The common difference is any term minus the previous one: $d = -1 - (-5) = -1 + 5 = 4$

So $a = -5$ and $d = 4$.

Here $a = 2$, $d = 5 - 2 = 3$, and $n = 12$.

Using $a_n = a + (n-1)d$: $a_{12} = 2 + (12 - 1)\times 3 = 2 + 11\times 3 = 2 + 33 = 35$

The 12th term is $35$.

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

$a + (n-1)d = 78$ $3 + (n-1)\times 5 = 78$ $(n-1)\times 5 = 75$ $n - 1 = 15 \implies n = 16$

So $78$ is the 16th term.

Using $a_n = a + (n-1)d$: $a_7 = a + 6d = 34 \quad (1)$ $a_{13} = a + 12d = 64 \quad (2)$

Subtract (1) from (2): $6d = 30 \implies d = 5$

Substitute in (1): $a + 6(5) = 34 \implies a = 34 - 30 = 4$

The AP is $4, 9, 14, 19, \dots$

Here $a = 11$, $d = 8 - 11 = -3$.

Suppose $a_n = 150$: $11 + (n-1)(-3) = 150$ $(n-1)(-3) = 139$ $n - 1 = -\frac{139}{3}$

This is not a positive integer, so $150$ is not a term of the AP. (Also note this AP decreases, so it never reaches 150.)

The two-digit multiples of 3 are $12, 15, 18, \dots, 99$.

This is an AP with $a = 12$, $d = 3$, last term $a_n = 99$.

$a + (n-1)d = 99$ $12 + (n-1)3 = 99$ $(n-1)3 = 87 \implies n - 1 = 29 \implies n = 30$

There are 30 two-digit numbers divisible by 3.

Here $a = 5$, $d = 4$, $n = 20$.

Using $S_n = \dfrac{n}{2}\big[2a + (n-1)d\big]$: $S_{20} = \frac{20}{2}\big[2(5) + (20-1)4\big]$ $= 10\big[10 + 76\big] = 10 \times 86 = 860$

The sum is $860$.

The first 15 multiples of 8 are $8, 16, 24, \dots$ up to $8 \times 15 = 120$.

So $a = 8$, $d = 8$, $n = 15$, last term $l = 120$.

Using $S_n = \dfrac{n}{2}(a + l)$: $S_{15} = \frac{15}{2}(8 + 120) = \frac{15}{2}\times 128 = 15 \times 64 = 960$

The sum is $960$.

Here $a = 9$, $d = 8$, $S_n = 636$.

$S_n = \frac{n}{2}\big[2a + (n-1)d\big] = 636$ $\frac{n}{2}\big[18 + (n-1)8\big] = 636$ $\frac{n}{2}\big[8n + 10\big] = 636$ $n(4n + 5) = 636$ $4n^2 + 5n - 636 = 0$

Solving: $n = \dfrac{-5 \pm \sqrt{25 + 10176}}{8} = \dfrac{-5 \pm 101}{8}$.

Taking the positive value: $n = \dfrac{96}{8} = 12$.

So 12 terms are required.

The $n$th term is $a_n = S_n - S_{n-1}$.

$S_n = 3n^2 + 5n$ $S_{n-1} = 3(n-1)^2 + 5(n-1) = 3(n^2 - 2n + 1) + 5n - 5 = 3n^2 - 6n + 3 + 5n - 5$ $S_{n-1} = 3n^2 - n - 2$

Therefore: $a_n = (3n^2 + 5n) - (3n^2 - n - 2) = 6n + 2$

The 10th term: $a_{10} = 6(10) + 2 = 62$

Length of a semicircle of radius $r$ is $\pi r$.

Lengths: $0.5\pi, 1.0\pi, 1.5\pi, \dots$ — an AP with $a = 0.5\pi$, $d = 0.5\pi$, $n = 13$.

Sum: $S_{13} = \frac{13}{2}\big[2(0.5\pi) + (13-1)(0.5\pi)\big]$ $= \frac{13}{2}\big[\pi + 6\pi\big] = \frac{13}{2}\times 7\pi = \frac{91\pi}{2}$

With $\pi = \tfrac{22}{7}$: $S_{13} = \frac{91}{2}\times \frac{22}{7} = \frac{91 \times 22}{14} = \frac{2002}{14} = 143 \text{ cm}$

Total length is $143$ cm.