If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Monjur Alahi

0 like 0 dislike
2 views
Share with your friends

2 Answers

Salim Ahmed Kawsar

Call

Sum of first n terms of A.P will be n2 

Explanation :
n1=7
Sn=49
we know that Sn = n/2{2a+(n-1)d}

thus,
Sn1=7/2(2a+6d)=49
    =14a+42d=98     ... (1) x17

n2=17
Sn2=289
Sn2=17/2(2a+16d)=289
     =34a+272d=578     ....(2) x7
from eq 1 & 2
238a +714d =1666    ....(3)
238a +1904d=4046    .....(4)

subtracting (4) -(3)
1190d=2380
    d=2
    a=1

sum of n terms will be  Sn = n/2{2a+(n-1)d}
Sn = n2 
 

0 like 0 dislike
2 views
Talk Doctor Online in Bissoy App
Salim Ahmed Kawsar

Call

given that `s_7 = 49 = 7^2 (s_n = n^2)`
`s_17= 289 = 17^2 `
now `s_n=n/2[2a+ (n-1)d]`
`s_7= 7/2[2a+(7-1)d]`
`49= 7/2[2a+6d]`
`7=a+3d` eqn(1)
now `S_17= 17/2[2a+(17-1)d]`
`289= 17/2[2a+ 16d]`
`a+8d=17`eqn(2)
subtracting eqns (1) & (2)
we get `5d=10`
`d=2`
puting it in eqn (1)
`a+3(2)=7`
`a=1`
`s_n= n/2[2a+(n-1)d]`
`s_n= n/2[2+(n-1)2]`
`=n/2[2+2n-2]`
`s_n=n^2`
answer

0 like 0 dislike
2 views
Talk Doctor Online in Bissoy App
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Let 1st term of AP be a Its common difference d = 3rd term - 2nd term = 18-14=4 So  a+d =14=> a=14-4=10 Sum of first 51 terms =51/2(2*10+(51-1)*4) = 51×110=5610

1 Answers 1 views
If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)?

Given sum up to n th term Sn= 4n-n^2 So   n th term wil be  tn = Sn - Sn-1= 4n -n^2-(4(n-1)-(n-1)^2 => tn= 4+2n -1=5-2n So sum of first 2 term S2= 4×2-2^2=4 t1 =...

1 Answers 1 views
After a deduction of 5% from a certain sum and 10% from the remainder, a sum of Rs 171 is left. The original sum was- ( The options are given below)

Please explain it.

1 Answers 1 views
After a deduction of 5% from a certain sum and 10% from the remainder, a sum of Rs 171 is left. The original sum was- ( The options are given below) 1 Answers 1 views
After a deduction of 5% from a certain sum and 10% from the remainder, a sum of Rs 171 is left. The original sum was- ( The options are given below) 1 Answers 1 views
Find the sum of the first k terms of a series whose n th term is 2an+b The n th term of the AP is given by 2an+b

a1=2a+b a2 =4a+b a3=6a+b Common difference = d=( 4a + b ) - ( 2a + b ) = 2a Therefore, sum of first k terms =k/2[(2a+(k-1)d]= k/2[(2(2a+b)+(k-1)2a]= k/2 x 2 (2a+b+k-a)=k(a+b+ak)

1 Answers 1 views
Find the sum of the first 2n terms of the following series. 12 - 22 + 32 - 42 + .....

We want to find 12 - 22 + 32 - 42 +..... to 2n terms = 1 - 4 + 9 - 16 + 25 ------------2n terms = (1 – 4) +(9...

1 Answers 1 views
Find the sum of n terms of an A.P.whose nth terms is given by an= 5 - 6n.

We have, an = 5-6n ⇒ a1 = 5 - 6 x 1 = -1 So, the given sequence is an A.P with first term a = a1 = -1 and last...

1 Answers 1 views
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and the 10th terms is 44. Find the first three terms of the AP.

Solution: a + 3d + a + 7d = 24 Or, 2a + 10d = 24 Similarly, 2a + 14d = 44 So, 44 – 24 = 4d Or, d = 5 2a + 10x5...

1 Answers 2 views
The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of the middle terms

Let the four consecutive numbers in AP be (a - 3d), (a - d), (a + d) and (a + 3d) So, according to the question. a-3d + a - d +...

1 Answers 1 views
X