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Question 1

What is the average of the 9 consecutive intergers starting with a + 2, if the average of 5 consecutive numbers is given as b?

A

b + 4

B

a + 4

C

b + 2

D

a + 2

Question 1 Explanation:

If take exmaple with numerical values. Avg. of 1 to 5 is 3. Hence if we take a = 1, b = 3. Now for interger starting with 1 +2 = 3, avg for 3,6,7,8,9,10,11,12,13 is Avg. is 7, hence avg. is b +4

Question 2

Assume there is an AP with 4 terms, the product of extremes is 88 and sum of means is 24, what is the third term in AP?

A

32

B

12

C

22

D

14

Question 2 Explanation:

When ever the terms are not given always assume like the terms cancel each other. In this case we will assume a -3d, a -d, a+ d, a +3d. Hence the sum of the means = a -d +a + d = 24 or a =12.
Product of extremes give (a -3d)(a +3d) = a^2 - 9d^2 = 88, solving this give d =2. Hence the terms are 6, 10, 14, 18

Question 3

How many 3 digit positive integers exists when divded by 9 leave remain 5 as remainder?

A

99

B

110

C

101

D

100

Question 3 Explanation:

The lowest 3 digit number leaving remainder 5 when divided by 9 is 104. The largest number is 995. Hence now we have sum of AP progression with 1st term as 104 and common difference 9. Number of terms can be found as 995 = 104 +(n-1)9 , n= 100

Question 4

How many numbers in the series -16, -12, -8, -4, 0, 4, 8, .... we should take so that sum is 72?

A

12

B

15

C

11

D

9

Question 4 Explanation:

Answer is 11, quite simple.... the negative term and positive term till 16 will cancel. This is a series of 4 as -16,-12,-8,-4,0,4,8,12,16, 20,24, 28. The last 3 adds to 72

Question 5

What is 51^{st} term of an AP starting with 5 with common difference of 10?

A

505

B

515

C

415

D

405

Question 5 Explanation:

nth term = a +(n-1)d = 5+ 50x10 = 505

Question 6

Sum of 5th and 11th term of an AP is 40. What is the sum of 15 terms of this AP?

A

250

B

300

C

200

D

150

Question 6 Explanation:

sum of 5th term and 11 term is (a+4d) +(a+10d) = 40. This is 2a+14d = 40, Sum of 15 terms (n/2)(2a + 14d). (15/2) x 40 = 300

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