Progression practice test
Find the 15th term of an arithmetic progression whose first term is 2 and the common difference is 3.
n th term of A.P = a +(n-1) *d
= 2+(15-1)*3 , = 2 + 42 = 44.
What is the sum of the first 15 terms of an A.P whose 11 th and 7 th terms are 5.25 and 3.25 respectively
a +10d = 5.25, a+6d = 3.25, 4d = 2, d = ½
a +5 = 5.25, a = 0.25 = ¼, s 15 = 15/2 ( 2 * ¼ + 14 * ½ )
= 15/2 (1/2 +14/2 ) = 15/2 *15/2 =225/ 4 = 56.25
If(12+22+32+…..+102)=385,then the value of (22+42+62+…+202) is :
22+42+62+…202=(1*2)2+(2*2)2+(2*3)2+…..+(2*10)2
=22*12+22*22+22*32+…22*102
=22[12+22+….102]
=4*385=1540
In an arithmetic series consisting of 51 terms, the sum of the first three terms is 65 and the sum of the middle three terms is 129. What is the first term and the common difference of the series?
Given a + (a + d) + (a + 2d)= 65
=> 3a + 3d =65…. (1)
The middle terms are 25th , 26th and 27th terms
=> a + 24d + a + 25d + a + 26d = 129
=> 3a + 75d
129 ….(2)
From (1) and (2) d= 8/9 =>a= 187/9
The sum of the first 100 numbers, 1 to 100 is divisible by
The sum of the first 100 natural numbers is given by
= [n * (n + 1) ]/ 2
= [100 * 101]/2
= 50 * 101
101 is an odd number and 50 are divisible by 2.
Hence, 50*101 will be divisible by 2.
How many terms are there in G.P 3,6,12,24,….,384?
Here a=3 and r=6/3=2.Let the number of terms be n
Then,tn=384=>arn-1=384
=>3*2n-1=384=>2n-1=128=27
=>n-1=7=>n=8
There for number of terms=8
Four angles of a quadrilateral are in G.P. Whose common ratio is an intiger. Two of the angles are acute while the other two are obtuse. The measure of the smallest angle of the quadrilateral is
Let the angles be a, ar, ar 2, ar 3.
Sum of the angles = a ( r 4- 1 ) /r -1 = a ( r 2 + 1 ) ( r + 1 ) = 360
a< 90 , and ar< 90, Therefore, a ( 1 + r ) < 180, or ( r 2 + 1 ) > 2
Therefore, r is not equal to 1. Trying for r = 2 we get a = 24 Therefore, The angles are 24, 48, 96 and 192.
How many numbers between 11 and 90 divisible by 7?
The required numbers are 14,21,28….84
This is an A.P with a=14 and d=(21-14)=7
Let it contain n terms
Then,Tn=84=>a+(n-1)d=84
=>14+(n-1)*7=84 or n=11
there for required number of terms=11
If Sn denotes the sum of the first n terms in an Arithmetic Progression and S1: S4 = 1: 10
Then the ratio of first term to fourth term is:
Use Sn = (n/2)[ 2a + (n-1)d] and Tn = a + (n – 1) d
S1/S4 = 1/10 = a/ (4/2) [2a + 3d]
6a = 6d or a = d
Therefore T1/T4 = a/ (a+ 3d) = a/4a = ¼
The sum of the three numbers in A.P is 21 and the product of their extremes is 45. Find the numbers.
Let the numbers are be a – d, a, a + d
Then a – d + a + a + d = 21
3a = 21
a = 7
and (a – d)(a + d) = 45
a2 – d2 = 45
d2 = 4
d = +2
Hence, the numbers are 5, 7 and 9 when d = 2 and 9, 7 and 5 when d = -2.
In both the cases numbers are the same.