Arithmetic Progression — Balbharati Solutions, AI Tutor & Practice

A sequence is an A.P. if the difference between consecutive terms is constant.

Compute differences:

• $7 - 3 = 4$
• $11 - 7 = 4$
• $15 - 11 = 4$
• $19 - 15 = 4$

Since the common difference is the same throughout, the sequence is an A.P.

First term $a = 3$ and common difference $d = 4$.

Given $a = 5$ and $d = -3$.

Each term is obtained by adding $d$ to the previous term.

• $t_1 = a = 5$
• $t_2 = 5 + (-3) = 2$
• $t_3 = 2 + (-3) = -1$
• $t_4 = -1 + (-3) = -4$

The A.P. is $5, 2, -1, -4, \ldots$

The nth term of an A.P. is given by: $t_n = a + (n-1)d$

Here $a = 7$, $d = 13 - 7 = 6$, and $n = 19$.

$t_{19} = 7 + (19 - 1)\times 6$ $t_{19} = 7 + 18 \times 6 = 7 + 108 = 115$

The 19th term is $\boxed{115}$.

Here $a = 3$, $d = 8 - 3 = 5$, and $t_n = 88$.

Using $t_n = a + (n-1)d$: $88 = 3 + (n-1)\times 5$ $88 - 3 = (n-1)\times 5$ $85 = 5(n-1)$ $n - 1 = 17 \implies n = 18$

So $88$ is the 18th term of the A.P.

Using $t_n = a + (n-1)d$:

$t_5 = a + 4d = 22$ ... (i)

$t_9 = a + 8d = 42$ ... (ii)

Subtract (i) from (ii): $(a + 8d) - (a + 4d) = 42 - 22$ $4d = 20 \implies d = 5$

Substitute $d = 5$ in (i): $a + 4(5) = 22 \implies a + 20 = 22 \implies a = 2$

First term $a = 2$ and common difference $d = 5$.

The sum of the first $n$ terms is: $S_n = \frac{n}{2}\left[2a + (n-1)d\right]$

Here $a = 4$, $d = 9 - 4 = 5$, $n = 25$.

$S_{25} = \frac{25}{2}\left[2(4) + (25-1)\times 5\right]$ $= \frac{25}{2}\left[8 + 120\right]$ $= \frac{25}{2}\times 128 = 25 \times 64 = 1600$

The sum of the first 25 terms is $\boxed{1600}$.

Numbers divisible by 7 between 1 and 200: $7, 14, 21, \ldots, 196$.

This is an A.P. with $a = 7$, $d = 7$, last term $l = 196$.

Find number of terms using $t_n = a + (n-1)d$: $196 = 7 + (n-1)\times 7$ $189 = 7(n-1) \implies n - 1 = 27 \implies n = 28$

Sum using $S_n = \frac{n}{2}(a + l)$: $S_{28} = \frac{28}{2}(7 + 196) = 14 \times 203 = 2842$

The required sum is $\boxed{2842}$.

The nth term can be found using: $t_n = S_n - S_{n-1}$

Given $S_n = 3n^2 + 5n$.

Compute $S_{10}$: $S_{10} = 3(10)^2 + 5(10) = 300 + 50 = 350$

Compute $S_{9}$: $S_{9} = 3(9)^2 + 5(9) = 243 + 45 = 288$

Therefore: $t_{10} = S_{10} - S_{9} = 350 - 288 = 62$

The 10th term is $\boxed{62}$.

Given $t_3 = 8$: $a + 2d = 8 \quad \text{...(i)}$

'The 7th term exceeds three times the 3rd term by 2' means: $t_7 = 3 \times t_3 + 2 = 3(8) + 2 = 26$

So: $a + 6d = 26 \quad \text{...(ii)}$

Subtract (i) from (ii): $4d = 18 \implies d = 4.5$

Substitute in (i): $a + 2(4.5) = 8 \implies a + 9 = 8 \implies a = -1$

First term $a = -1$, common difference $d = 4.5$.

Monthly savings form an A.P.: $100, 120, 140, \ldots$

Here $a = 100$, $d = 20$, $n = 12$.

Use the sum formula: $S_n = \frac{n}{2}\left[2a + (n-1)d\right]$ $S_{12} = \frac{12}{2}\left[2(100) + (12-1)\times 20\right]$ $= 6\left[200 + 220\right]$ $= 6 \times 420 = 2520$

The total amount saved in 12 months is $\boxed{₹2520}$.

Here $a = 24$, $d = 21 - 24 = -3$, and $S_n = 78$.

$S_n = \frac{n}{2}\left[2a + (n-1)d\right]$ $78 = \frac{n}{2}\left[48 + (n-1)(-3)\right]$ $156 = n\left[48 - 3n + 3\right]$ $156 = n(51 - 3n)$ $156 = 51n - 3n^2$ $3n^2 - 51n + 156 = 0$ $n^2 - 17n + 52 = 0$ $(n - 4)(n - 13) = 0$ $n = 4 \text{ or } n = 13$

Both values are valid because the common difference is negative — later terms become negative, so the sum returns to 78.

So 4 terms or 13 terms must be added.

Let the three consecutive terms be $a - d$, $a$, $a + d$.

Sum condition: $(a-d) + a + (a+d) = 27$ $3a = 27 \implies a = 9$

Product condition: $(a-d)(a)(a+d) = 504$ $9(a^2 - d^2) = 504$ $9(81 - d^2) = 504$ $81 - d^2 = 56$ $d^2 = 25 \implies d = \pm 5$

Taking $d = 5$: the terms are $4, 9, 14$.

(Taking $d = -5$ gives $14, 9, 4$ — the same numbers in reverse.)

The three terms are $\boxed{4, 9, 14}$.