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Progression

Practice aptitude questions

Q 1 - What number of term arrives in the A.P. 7, 11, 15..... 139?

Solution

Here a = 7 and d = (11-7) = 4
Let there be n terms in the given A.P. then,
Tn = 139 ⇒a+ (n-1) d =139
⇒ 7+ (n-1)*4 = 139 ⇒ (n-1) *4 = 132 ⇒ (n-1) = 33
⇒ n = 34

Q 2 - In the event that the fourth term of a number juggling movement is 14 and its twelfth term is 70, then its first term is:

Solution

Let the first term of the A.P. be a and common difference be d. then,
a+ 3d = 14 ...(i)    a+11d = 70 ...(ii)
On subtracting (i) from (ii), we get 8d =56    d = 7
Putting d = 7 in (i), we get a+3*7=14 ⇒ a= (14-21) = -7   ∴ First term = -7

Q 3 - What number of products of arrive between the whole numbers 15 and 105, both comprehensive?

Solution

Requisite no. are 15, 18, 21, 24, 105.
Here a = 15 and d = (18-15) = 3
Let their number be n. Then,
a + (n-1) d =105 ⇒ 15+ (n-1)*3 = 105
⇒ a+ (n-1) d = 105 ⇒ 15+ (n-1)*3 = 105
⇒ (n-1)*3 = 90   ⇒ n-1 =30 ⇒ n = 31

Q 4 - The total of the all odd number somewhere around 100 and 200 is:

Solution

 Requisite sum = 101 +103 + 105+...+199.
This is an A.P. in which a = 101, d =2 and L= 199.
A + (n-1) d =199 ⇒ 101 + (n-1) *2 = 199
⇒ (n-1) * 2 = 98 ⇒ (n-1) = 49 ⇒ n = 50.
∴ Sum = n/2 * (a+L) = 50/2 * (101 +199) = (50*150) =7500.

Q 5 - The total of every single common number from 75 to 97 is:

Solution

Sum = 75+76+77+...+97.
Here a =75, d = (76-75) =1
Let the number of terms be n. Then,
A+ (n-1) d =97⇒ 75 + (n-1)*1 =97 ⇒ (n-1) = 22 ⇒ n= 23.
∴ Sum = 23/2 (75 + 97) = (23/2 *172) = (23 *86 ) = 1978.

Q 6 - In the event that the fourth and ninth term of a G.P. is 54 and 13122 separately, there its second term is:

Solution

Let its 1st term be a and common ratio r. Then,
ar3 = 54 and ar⁸ = 13122
∴ ar⁸/ ar3 = 13122/54 ⇒ r⁵ =243 = 3⁵ =r = 3
∴ a* 33 = 54 ⇒ a*27 = 54 = > a =2
2nd term = ar = (2*3) =6

Q 7 - Three numbers are in G.P. Their whole is 28 and item is 512. The numbers are:

Solution

Let the number be a/r, a, ar.
Then, a/r* a*ar = 512 ⇒ a3 =83 ⇒ a =8
8/r+8+8r =28 ⇒ 8/r +8r =20 ⇒ 2/r+2r =5
⇒ 2r2-5r+2 =0 ⇒ 2r2-4r-r+2 =0
⇒ 2r (r-2)- (r-2)=0   ⇒ (r-2)(2r-1) =0
⇒r = 2 or r = 1/2
Numbers are 4,8, 16.

Q 8 - A few buys National Savings Certificates each year whose worth surpasses the earlier years buy by Rs 400. Following 8 years, she finds that she has obtained declarations whose aggregate face worth is Rs 48000. What is the face estimation of the Certificates acquired by her in the first years?

Solution

Let the required value be Rs a.
Also d= 400, n = 8 and Sn = 48000.
Sn   = n/2 [2a + (n -1) d] ⇒ 8/2 *[2a + 7 *400] = 48000
⇒ 2a + 2800 = 12000⇒ 2a=9200⇒ a = 4600

Q 9 - (142 +152 +......+302) =?

Solution

We know that (12+22+32+... +a2) = {n(n+1)(2n+1)}/6
Given Exp. = (12+22+...+132+142+??+302)-(12+22+...+132)
= (30*31*61)/6 - (13*14*27)/6 = (9455- 819) = 8636

Q10 Find the position of 62 in the following series 2, 5, 8, ....?

Solution

It is an A.P. series with a = 2, d = 3
 
Let 62 be the nth term.
 
 Tn = 62
 
But we know that Tn = a + (n � 1) d
 
 
  62 = 2 + (n � 1) 3      
 
or 62 = 2 + 3n � 3
 
or 62 � 2 + 3 = 3n
 
or 63 = 3n
 
or� 3n = 63
 
or n = 21
 
62 is the 21st term