If “a” is the first term and ℓ is the last term (nth term), then the sum of all the term of this sequence is given by:
By formula
Sn = n/2(a+l)
Find the sum of first hundred even natural numbers which are divisible by 5.
Even natural numbers which are divisible by 5 are 10,20.30...
Difference between the consecutive terms is same
∴ They form an AP whose first term (a) = 10,
common on difference (d) = 10,
and number of terms (n) = 100
∴ The sum of first hundred even natural numbers which are divisible by 5 is 50500.
If the 10 times of the 10th term of an AP is equal to 15 times to the 15th term, then the 25th term is:
10 x 10th term = 15 x 15th term
let a is the first term and d is the common difference .
10(a+9d) = 15(a+14d)
5a + 120d = 0
a + 24d = 0
now , 25th term = a + (251)d
= a+24d = 0
hence 25th term = 0
Find the missing number. 1, 4, 9, 16, 25, 36, 49, (....)
The series is 1^{2}, 2^{2}, 3^{2}, 4^{2}, 5^{2}, 6^{2}, 7^{2}, ...
Hence, next term = 8^{2} = 64
The first negative term of the A.P.62,57,52…. is the
a = 62 d = 57  62 = 5
tn = a + (n1)d
= 62 + (n1)(5)
= 62  5n + 5
= 67  5n
From the options, we take '14'
= 67  5(14)
= 67  70
= 3 (The first negative term will be at the 14th term)
How many terms of the series 24,20,16,…are required so that their sum is 72?
The series is 24,20,16...in AP
so a = 1st term=24 & common difference = 2024 = 4
sum = 72 is of say n terms
then 72 = n/2[2*24+(n1)*4] = n/2[484n+4] = n/2[524n]
or 144 = 52n4n^{2}
or 4n^{2}52n+144 = 0
or n^{2}13n+36 = 0
or n^{2}9n4n+36 = 0
or n(n9)4(n9) = 0
or (n9)(n4) = 0
or n = 9 or 4
Roots of quadratic equation x^{2} – 3x = 0 , will be
Given x ^{2}  3x = 0
Factor x out in the expression on the left.
x (x  3) = 0
For the product x (x  3) to be equal to zero we nedd to have
x = 0 or x  3 = 0
Solve the above simple equations to obtain the solutions.
x = 0
or
x = 3
The number of terms in the sequence 17, 10, 3,…., 144 is:
Three terms in A.P. are such that their sum is 45. What is the middle term?
Let the three numbers be ad, a, a+d
ATQ,
a  d + a + a + 2d = 45
3a = 45
a = 15
Middle term is 15
The terms of an A.P. are doubled, then the resulting sequence is
The general form of an AP is a, a+d, a+2d,.....
where a is the first term and d is the common difference
If we double the terms, the new sequence would be
A , a+2d, a+4d,......
We can observe that this sequence is also an AP
First term is a
Common difference is 2d nth term= 2a+(n1)2d
= 2[a+(n1)d]
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