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QUESTION: 1

If “a” is the first term and ℓ is the last term (nth term), then the sum of all the term of this sequence is given by:

Solution:

By formula

Sn = n/2(a+l)

QUESTION: 2

Find the sum of first hundred even natural numbers which are divisible by 5.

Solution:

Even natural numbers which are divisible by 5 are 10,20.30...

Difference between the consecutive terms is same

∴ They form an AP whose first term (a) = 10,

common on difference (d) = 10,

and number of terms (n) = 100

∴ The sum of first hundred even natural numbers which are divisible by 5 is 50500.

QUESTION: 3

If the 10 times of the 10th term of an AP is equal to 15 times to the 15th term, then the 25th term is:

Solution:

10 x 10th term = 15 x 15th term

let a is the first term and d is the common difference .

10(a+9d) = 15(a+14d)

5a + 120d = 0

a + 24d = 0

now , 25th term = a + (25-1)d

= a+24d = 0

hence 25th term = 0

QUESTION: 4

Find the missing number. 1, 4, 9, 16, 25, 36, 49, (....)

Solution:

The series is 1^{2}, 2^{2}, 3^{2}, 4^{2}, 5^{2}, 6^{2}, 7^{2}, ...

Hence, next term = 8^{2} = 64

QUESTION: 5

The first negative term of the A.P.62,57,52…. is the

Solution:

a = 62 d = 57 - 62 = -5

tn = a + (n-1)d

= 62 + (n-1)(-5)

= 62 - 5n + 5

= 67 - 5n

From the options, we take '14'

= 67 - 5(14)

= 67 - 70

= -3 (The first negative term will be at the 14th term)

QUESTION: 6

How many terms of the series 24,20,16,…are required so that their sum is 72?

Solution:

The series is 24,20,16...in AP

so a = 1st term=24 & common difference = 20-24 = -4

sum = 72 is of say n terms

then 72 = n/2[2*24+(n-1)*-4] = n/2[48-4n+4] = n/2[52-4n]

or 144 = 52n-4n^{2}

or 4n^{2}-52n+144 = 0

or n^{2}-13n+36 = 0

or n^{2}-9n-4n+36 = 0

or n(n-9)-4(n-9) = 0

or (n-9)(n-4) = 0

or n = 9 or 4

QUESTION: 7

Roots of quadratic equation *x*^{2} – 3*x* = 0 , will be

Solution:

Given x ^{2} - 3x = 0

Factor x out in the expression on the left.

x (x - 3) = 0

For the product x (x - 3) to be equal to zero we nedd to have

x = 0 or x - 3 = 0

Solve the above simple equations to obtain the solutions.

x = 0

or

x = 3

QUESTION: 8

The number of terms in the sequence -17, -10, -3,…., 144 is:

Solution:

QUESTION: 9

Three terms in A.P. are such that their sum is 45. What is the middle term?

Solution:

Let the three numbers be a-d, a, a+d

ATQ,

a - d + a + a + 2d = 45

3a = 45

a = 15

Middle term is 15

QUESTION: 10

The terms of an A.P. are doubled, then the resulting sequence is

Solution:

The general form of an AP is a, a+d, a+2d,.....

where a is the first term and d is the common difference

If we double the terms, the new sequence would be

A , a+2d, a+4d,......

We can observe that this sequence is also an AP

First term is a

Common difference is 2d nth term= 2a+(n-1)2d

= 2[a+(n-1)d]

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