CBSE Class 10 Answered
Question 1)
Find the sum of first 40 positive integers divisible by 6.
Question 2)
Find the sum of n terms of the series (4-1/n)+4-2/n)+(4-3/n)+........
Question 3)
If the sum of the first n terms of an AP is 1/2(3n²+7n), then find its nth term .Hence write the 20 th term .
Asked by arindeep.singh | 22 Jul, 2020, 06:18: PM
Expert Answer
(1)
We know that the first 40 positive integers divisible by 6 are 6,12,18,....
This is an AP with a = 6 and d = 6
S40= 20[2(6) +(40-1)6] =20[12+234] =4920
(2)
(4-1/n)+(4-2/n)+(4-3/n)+.....+(4-n/n)
= (4+4+...+upto n terms) - 1/n(1+2+3+...+n)
= 4n - 1/n[n(n+1)/2]
= 4n - (n+1)/2
= (8n - n - 1)/2
= (7n-1)/2
(3)
Given: Sn = 1/2 (3n2+7n)
Take n=1, S1 = 1/2 (3+7) = 5
Take n=2, S2 = 1/2 (3x4 + 7x2) = 26/2 = 13
We know, S1=a1=5
S2=a1+a2=13
S2-S1=a1+a2-a1
13-5=a2
Therefore, a2=8
Also, d = a2-a1
d = 8-5 = 3
nth term of AP = an = 5+(n-1)3
an= 2+3n
Therefore 20th term is
a20= 2+3(20)=62
Hence 20th term is 62.
Given: Sn = 1/2 (3n2+7n)
Take n=1, S1 = 1/2 (3+7) = 5
Take n=2, S2 = 1/2 (3x4 + 7x2) = 26/2 = 13
We know, S1=a1=5
S2=a1+a2=13
S2-S1=a1+a2-a1
13-5=a2
Therefore, a2=8
Also, d = a2-a1
d = 8-5 = 3
nth term of AP = an = 5+(n-1)3
an= 2+3n
Therefore 20th term is
a20= 2+3(20)=62
Hence 20th term is 62.
We know that the first 40 positive integers divisible by 6 are 6,12,18,....
This is an AP with a = 6 and d = 6
S40= 20[2(6) +(40-1)6] =20[12+234] =4920
Answered by Renu Varma | 23 Jul, 2020, 01:00: PM
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