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Choose the most appropriate option ( a ), ( b ) , ( c ) or (d)
The nth element of the sequence 1, 3, 5, 7,….…..Is
Let a be the first term and d be the common difference.
Given series is in AP.
a = 1
d = 3  1 = 2
We know that nth term of an Ap an = a + (n  1) * d
= 1 + (n  1) * 2
= 1 + 2n  2
= 2n  1.
The nth element of the sequence –1, 2, –4, 8 ….. is
The arithmetic mean between a and 10 is 30, the value of ‘a’ should be
–5, 25, –125 , 625, ….. can be written as
The first three terms of sequence when nth term tn is n^{2} – 2n are
Which term of the progression –1, –3, –5, …. Is –39
The value of x such that 8x + 4, 6x – 2, 2x + 7 will form an AP is
The mth term of an A. P. is n and nth term is m. The r th term of it is
An infinite GP has first term x and sum 5, then x belongs to
We know that, the sum of infinite terms of GP is
The nth term of the series whose sum to n terms is 5n^{2} + 2n is
It is given that the sum of n terms is
The 20^{th} term of the progression 1, 4, 7, 10.................is
The last term of the series 5, 7, 9,….. to 21 terms is
The last term of the A.P. 0.6, 1.2, 1.8,… to 13 terms is
The sum of the series 9, 5, 1,…. to 100 terms is
The two arithmetic means between –6 and 14 is
Let the terms be – 6, a, b 14
a = – 6
T4 = a + 3d = 14
– 6 + 3d = 14
3d = 20
d = 20/3
a = – 6 + 20/3 = 2/3
b = 2/3 + 20/3 = 22/3
The sum of three integers in AP is 15 and their product is 80. The integers are
let a  d , a , a + d are three terms of an AP
according to the problem given,
sum of the terms = 15
a  d + a + a + d = 15
3a = 15
a = 15 / 3
a = 5
product = 80
( a  d ) a ( a + d ) = 80
( a²  d² ) a = 80
( 5²  d² ) 5 = 80
25  d² = 80 /5
25  d² = 16
 d² =  9
d² = 3²
d = ± 3
Therefore,
a = 5 , d = ±3
required 3 terms are
a  d = 5  3 = 2
a = 5
a+ d = 5 + 3 = 8
( 2 , 5 , 8 ) or ( 8 , 5 , 2 )
The sum of n terms of an AP is 3n^{2} + 5n. A.P. is
The sum of n terms of an A.P. = 3n^{2}+5n
then sum of n1 terms= 3(n1)^{2}+5(n1)
So nth term of A.P will be Tn=3n^{2}+5n 3(n1)^{2}5(n1)= 3n^{2}+5n3n^{2}3+6n5n+5=6n+2
in this way by putinng values of n= 1,2,3,4,5,6,7,_______n1, n
we will get the AP as 8,14,20,26,32,38,44,____
The number of numbers between 74 and 25556 divisible by 5 is
The pth term of an AP is (3p – 1)/6. The sum of the first n terms of the AP is
The arithmetic mean between 33 and 77 is
The 4 arithmetic means between –2 and 23 are
a=2 n=6
given : t6= 23
23= a+5d
23= 2+5d
d= 25/5
d = 5
t2= a+d= 2+5=3
t3= 8
t4= 13
t5= 18
The first term of an A.P is 14 and the sums of the first five terms and the first ten terms are equal is magnitude but opposite in sign. The 3^{rd} term of the AP is
The sum of a certain number of terms of an AP series –8, –6, –4, …… is 52. The number of terms is
The 1^{st} and the last term of an AP are –4 and 146. The sum of the terms is 7171. The number of terms is
The sum of the series 3 ½ + 7 + 10 ½ + 14 + …. To 17 terms is
The 7^{th} term of the series 6, 12, 24,……is
t_{8} of the series 6, 12, 24,…is
t_{12} of the series –128, 64, –32, ….is
The 4^{th} term of the series 0.04, 0.2, 1, … is
The last term of the series 1, 2, 4,…. to 10 terms is
The last term of the series 1, –3, 9, –27 up to 7 terms is
The last term of the series x^{2}, x, 1, …. to 31 terms is
The sum of the series –2, 6, –18, …. To 7 terms is
The sum of the series 24, 3, 8, 1, 2, 7,… to 8 terms is
Sum of series 24,3,8,1,2,7..to 8th term is:
Take the 1st and 2nd terms, 24,3, put them together, 243 = 3^5.
Take the 3rd and 4th terms, 8,1, put them together, 81 = 3^4.
Take the 5th and 6th terms, 2,7, put them together, 27 = 3^3.
Take the 7th and 8th terms, 0,9, put them together, 09 = 3^2.
Sum of 8 terms = 24+3+8+1+2+7+0+9 = 54
The sum of the series to 18 terms is
The second term of a G P is 24 and the fifth term is 81. The series is
The sum of 3 numbers of a G P is 39 and their product is 729. The numbers are
In a G. P, the product of the first three terms 27/8. The middle term is
If you save 1 paise today, 2 paise the next day 4 paise the succeeding day and so on, then your total savings in two weeks will be
If we save 1 paise today,2 paise the next day,4 pause the succeeding day and so on then you total savings in two weeks the sequence should be Geometric progression .
1, ,2,4, 8, 16, .....,14 terms
First term (a) = 1,
Sum of n terms of the series 4 + 44 + 444 + … is
S=4+44+444+n terms
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