## Find the sum of the odd numbers between 0 and 50.

[question] Question. Find the sum of the odd numbers between 0 and 50. [/question] [solution] Solution: 1, 3, 5, 7 ..., 49 a = 1, d = 2 $\ell=\mathrm{t}_{\mathrm{n}}=49$ $\Rightarrow a+(n-1) d=49$ $\Rightarrow 1+(n-1)(2)=49$ $\Rightarrow 1+2 n-2=49$ $\Rightarrow 2 n=50$ or $n=25$ The sum $=\frac{25}{2}\{a+\ell\}=\frac{25}{2}\{1+49)$ $=\frac{25}{2} \times 50=625$ [/solution]...

## Find the sum of the first 15 multiples of 8.

[question] Question. Find the sum of the first 15 multiples of 8. [/question] [solution] Solution: The multiples of 8 are 8, 16, 24, 3... These are in an A.P., having first term as 8 and common difference as 8. Therefore, a = 8 d = 8 $S_{15}=?$ $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $=\frac{15}{2}[2(8)+(15-1) 8]$ $=\frac{15}{2}[16+14(8)]$ $=\frac{15}{2}(16+112)$ $=\frac{15(128)}{2}=15 \times 64$ $=960$ [/solution]...

## Find the sum of the first 40 positive integers divisible by 6.

[question] Question. Find the sum of the first 40 positive integers divisible by 6. [/question] [solution] Solution: The positive integers that are divisible by 6 are 6, 12, 18, 24 .... It can be observed that these are making an A.P. whose first term is 6 and common difference is 6. a = 6 d = 6 $\mathrm{S}_{40}=?$ $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $\mathrm{S}_{40}=\frac{40}{2}[2(6)+(40-1) 6]$ = 20[12 + (39) (6)] = 20(12 + 234) = 20 × 246 = 4920 [/solution]...

## Find the sum of first 22 terms of an AP in which d = 7

[question] Question. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149. [/question] [solution] Solution: $d=7$ $a_{22}=149$ \mathrm{S}_{22}=? $a_{22}=a+(22-1) d$ 149 = a + 21 × 7 149 = a + 147 a = 2 $S_{n}=\frac{n}{2}\left(a+a_{n}\right)=\frac{22}{2}(2+149)=11(151)=1661$ [/solution]...

## The first and the last term of an AP are 17 and 350 respectively

[question] Question. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum? [/question] [solution] Solution: Given that, $a=17$ $\ell=350$ d = 9 Let there be n terms in the A.P. $\ell=a+(n-1) d$ 350 = 17 + (n – 1)9 333 = (n – 1)9 (n – 1) = 37 n = 38 $\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}(\mathrm{a}+\ell)$ $\Rightarrow S_{n}=\frac{38}{2}(17+350)=19(367)=6973$ Thus, this A.P. contains 38 terms and the ...

## Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs. 200 each year.

[question] Question. Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs. 200 each year. In which year did his income reach Rs. 7000 ? [/question] [solution] Solution: It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs. 200. Therefore, the salaries of each year after 1995 are 5000, 5200, 5400, ..... Here, a = 5000 d = 200 Let after $\mathrm{n}^{\text {th }}$ year, his sal...

## Which of the following are APs ?

[question] Question. Which of the following are APs ? If they form an AP, find the common difference d and write three more terms. (i) $2,4,8,16, \ldots$ (ii) $2, \frac{5}{2}, 3, \frac{7}{2}, \ldots$ (iii) $-1.2,-3.2,-5.2,-7.2, \ldots$ (iv) $-10,-6,-2,2, \ldots \ldots$ (v) $3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2}, \ldots$ (vi) $0.2,0.22,0.222,0.2222, \ldots .$ (vii) $\quad 0,-4,-8,-12, \ldots .$ (viii) $-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots$ (ix) $1,3,9,27, \ldots \ldots$ ...

## In which of the following situations,

[question] Question. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why? (i) The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km. (ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time. (iii) The cost of digging a well after every metre of digging, when it costs Rs. 150 for the first metre and rises by Rs. 50 for each s...