If the sum of seven terms of an AP is 49 and that of 17 terms is 289 find the sum of nth term.?

Question

Answer

Here, given that:-

S7=49 and S17=289

now,
Sn=n/2{2a+(n-1)d}
S7=7/2{2a+(7-1)d
49=7/2(2a+6d)
49/7=(a+3d)
7=a+3d. ----(1)

similarly,

S17=17/2{2a+(17-1)d}
289=17/2(2a+16d)
289/17=a+8d
17=a+8d. ----(2)

subracting (1) from (2), we get

17-7=a+8d-a-3d
10=5d
2=d

putting value of d in (1),

7=a+3×2
7-6=a
1=a

now, we have
a=1, d=2, n=n

so,
Sn=n/2{2×1+(n-1)2}
Sn=n(1+n-1)
Sn=n×n
Sn=n^2

So, the sum of nth term of given AP is n^2.

Thank you.. :-)

Answer

Answer

Problem:
The sum of seven terms of an AP is 49 and that of 17 terms is 289. Find the sum of nth term?

Solution:

Step 1: Finding the common difference(d) of the AP:
Let the first term of the AP be a and the common difference be d.
Then, the sum of seven terms of the AP is given by:
S7 = 7/2[2a + (7-1)d]
49 = 7/2[2a + 6d]
2a + 6d = 14 -------------(1)

Similarly, the sum of 17 terms of the AP is given by:
S17 = 17/2[2a + (17-1)d]
289 = 17/2[2a + 16d]
2a + 16d = 34 --------------(2)

Step 2: Solving the equations (1) and (2) to find the values of a and d:
Multiplying equation (1) by 2 and subtracting it from equation (2), we get:
2a + 16d - (4a + 12d) = 34 - 28
-2a + 4d = 3
a = (3-4d)/(-2) ---------------(3)

Substituting the value of a from equation (3) into equation (1), we get:
2[(3-4d)/(-2)] + 6d = 14
-3 + 6d = 14
d = 17/6

Substituting the value of d into equation (3), we get:
a = -1/2

Therefore, the AP is: -1/2, 2/3, 5/6, 7/3, 11/2, 23/3, ...

Step 3: Finding the sum of nth term of the AP:
The nth term of the AP is given by:
an = a + (n-1)d

The sum of n terms of the AP is given by:
Sn = n/2[2a + (n-1)d]

Substituting the values of a and d, we get:
an = -1/2 + (n-1)(17/6)
an = (17n-23)/6

Substituting the value of an into the formula of Sn, we get:
Sn = n/2[-1 + (17n-23)/6]

Therefore, the sum of nth term of the AP is (-n^2 + 17n + 23)/2.

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