Advertisements
Advertisements
प्रश्न
If Sn denotes the sum of first n terms of an AP, prove that S12 = 3(S8 – S4)
Advertisements
उत्तर
Sum of n terms of an AP,
∵ Sn = `n/2[2a + (n - 1)d]` ...(i)
∴ S8 = `8/2[2a + (8 - 1)d]`
= 4(2a + 7d)
= 8a + 28d
And S4 = `4/2[2a + (4 - 1)d]`
= 2(2a + 3d)
= 4a + 6d
Now, S8 – S4
= 8a + 28d – 4a – 6d
= 4a + 22d ...(ii)
And S12 = `12/2[2a + (12 - 1)d]`
= 6(2a + 11d)
= 3(4a + 22d)
= 3(S8 – S4) ...[From equation (ii)]
∴ S12 = 3(S8 – S4)
Hence proved.
APPEARS IN
संबंधित प्रश्न
The sum of three numbers in A.P. is –3, and their product is 8. Find the numbers
Find the sum of the following APs:
2, 7, 12, ..., to 10 terms.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
The 19th term of an AP is equal to 3 times its 6th term. If its 9th term is 19, find the AP.
How many two-digit number are divisible by 6?
Find the sum of all natural numbers between 200 and 400 which are divisible by 7.
If first term of an A.P. is a, second term is b and last term is c, then show that sum of all terms is \[\frac{\left( a + c \right) \left( b + c - 2a \right)}{2\left( b - a \right)}\].
What is the sum of first 10 terms of the A. P. 15,10,5,........?
The sum of third and seventh term of an A. P. is 6 and their product is 8. Find the first term and the common difference of the A. P.
A man is employed to count Rs 10710. He counts at the rate of Rs 180 per minute for half an hour. After this he counts at the rate of Rs 3 less every minute than the preceding minute. Find the time taken by him to count the entire amount.
The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be
The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its
Q.13
Q.18
The sum of the first three terms of an Arithmetic Progression (A.P.) is 42 and the product of the first and third term is 52. Find the first term and the common difference.
Which term of the AP 3, 15, 27, 39, ...... will be 120 more than its 21st term?
Find the sum of natural numbers between 1 to 140, which are divisible by 4.
Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136
Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
`square` = (n – 1) × 4
n = `square`
Now,
Sn = `"n"/2["a" + "t"_"n"]`
Sn = 17 × `square`
Sn = `square`
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
A merchant borrows ₹ 1000 and agrees to repay its interest ₹ 140 with principal in 12 monthly instalments. Each instalment being less than the preceding one by ₹ 10. Find the amount of the first instalment
If ₹ 3900 will have to be repaid in 12 monthly instalments such that each instalment being more than the preceding one by ₹ 10, then find the amount of the first and last instalment
Find the value of a25 – a15 for the AP: 6, 9, 12, 15, ………..
