Advertisements
Advertisements
Questions
In an AP given a = 3, n = 8, Sn = 192, find d.
Let there be an A.P. with the first term 'a', common difference 'd'. If an a denotes in nth term and Sn the sum of first n terms, find.
d, if a = 3, n = 8 and Sn = 192
Advertisements
Solution 1
Given that, a = 3, n = 8, Sn = 192
`S_n = n/2 [2a+(n-1)d]`
`192 = 8/2[2xx3+(8-1)d]`
192 = 4 [6 + 7d]
48 = 6 + 7d
42 = 7d
`d = 42/7`
d = 6
Solution 2
Here, we have an A.P. whose first term (a), the sum of first n terms (Sn) and the number of terms (n) are given. We need to find the common difference (d).
Here,
First term (a) = 3
Sum of n terms (Sn) = 192
Number of terms (n) = 8
So here we will find the value of n using the formula, an = a + (a - 1)d
So, to find the common difference of this A.P., we use the following formula for the sum of n terms of an A.P
`S_n = n/2 [2a + (n -1)d]`
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
So, using the formula for n = 8, we get,
`S_8 = 8/2 [2(3) + (8 - 1)(d)]`
192 = 4[6 + (7) (d)]
192 = 24 + 28d
28d = 192 - 24
Further solving for d
`d = 168/28`
d = 6
Therefore, the common difference of the given A.P. is d = 6.
APPEARS IN
RELATED QUESTIONS
In an AP, given a = 7, a13 = 35, find d and S13.
In an AP Given a12 = 37, d = 3, find a and S12.
Show that a1, a2,..., an... form an AP where an is defined as below:
an = 3 + 4n
Also, find the sum of the first 15 terms.
Find the sum of the following arithmetic progressions:
41, 36, 31, ... to 12 terms
Find the sum of the first 11 terms of the A.P : 2, 6, 10, 14, ...
How many two-digits numbers are divisible by 3?
Find the sum of first n even natural numbers.
The next term of the A.P. \[\sqrt{7}, \sqrt{28}, \sqrt{63}\] is ______.
Which term of the sequence 114, 109, 104, ... is the first negative term?
The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its
