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**Question 1:** Find the following sum>

1/(2^{2} –1) +1/(4^{2} –1) + 1/(6^{2} –1) + …. +1/(20^{2} –1)**(a)** 9/10**(b)** 10/11**(c)** 19/21**(d)** 10/21

**Answer (d)**

**Question 2:** Two men X and Y started working for a certain company at similar jobs on January 1, 1950. X asked for an initial salary of Rs. 300 with an annual increment of Rs. 30. Y asked for an initial salary of Rs. 200 with a rise of Rs. 15 every six months. Assume that the arrangements remained unaltered till December 31, 1959. Salary is paid on the last day of the month. What is the total amount paid to them as salary during the period?**(a)** Rs. 93,300**(b)** Rs. 93,200**(c)** Rs. 93,100**(d)** None of these

**Answer (a)**

The salary of X in the first year is Rs 300 and then it increases Rs 30 annually.

Hence, the total salary of X is given by:

X = 12 x (300 + 330 + 360 + 390 + 420 + 450 + 480 + 510 + 540 + 570)

[∵ sum of AP]

= 60 x 870 = Rs. 52,200

Similarly, the initial salary of Y was Rs 200 and it increased by Rs 15 every six months. So. the total salary of Y is given by

Y = 6 x [200+215 +230 + 245 +260 ........20 terms]

= 6 x 10 x [2 x 200-19 x 15] [sumofA.P.]

= 60 x [400 + 285] = Rs. 41.100

Hence, the total salary-paid = 52200 + 41100 = Rs 93300.

**Question 3: **Let S denote the infinite sum 2 + 5*x* + 9*x*^{2} + 14*x*^{3} + 20*x*^{4} + …, where | *x *| < 1, then *S* equals**(a)** {(2-x)/(1-x)^{3}}**(b)** {(2-x)/(1+x)^{3}}**(c)** {(2+x)/(1-x)^{3}}**(d)** {(2+x)/(1+x)^{3}}

**Answer (a)**

We will solve this question by using the given options.

In the first option we have,

Using the expansion for (1 - x)^{-3} = 1 + 3x + 6x^{2} + 10x^{3} + .....

= 2 + 5x+9x^{2} + 14x^{3} + ....

Which is equal to the given sum. Hence option 1 is the answer.

**Question 4:** The infinite sum 1 + (4/7) + (9/7^{2}) + (16/7^{3}) + (25/7^{4}) + …. equals**(a)** 27/14**(b)** 21/13**(c)** 49/27**(d)** 256/147

**Answer (c)**

We have .......(1)

.......(2)

Subtracting (2) from (1).

.......(3)

Multiplying (3) by 1 7, we obtain equation 4:

.......(4)

Subtracting (4) from (3).

**Question 5: Consider the set S = (1, 2, 3, …, 1000}. How man y arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least 3 elements?(a) 3(b) 4(c) 6(d) 7(e) 8 **

**Answer (d)**

Let number of terms in the arithmetic progression be n, then

1000 = 1 + (n–1) d

⇒ (n–1) d = 999

⇒ n – 1 = 999/d

Since n is an integer, so n – 1 is also an integer. This means that ‘d’ is a factor of 999.

Now 999 = 33 × 37. so the total factors of 999 are 4 × 2 = 8.

Out of these 8 factors one factor is 999 and we will reject it as in that case there will be only two terms in the

A.P. i.e. 1 and 1000, which is not possible.

Hence, ‘d’ can take 7 different values.

So, in total, 7 A.Ps. are possible.

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