Advertisements
Advertisements
प्रश्न
Sum of 1 to n natural number is 45, then find the value of n.
Advertisements
उत्तर
Natural numbers are 1, 2, 3 ......... n in AP.
Where, a = 1, d = 1
Sn = 45
∴ Sn = `n/2 [2a + (n - 1)d]`
⇒ 45 = `n/2 [2 xx 1 + (n - 1) xx 1]`
⇒ 90 = n[2 + n – 1]
∴ 90 = n[n + 1]
∴ n2 + n – 90 = 0
⇒ n2 + 10n – 9n – 90 = 0
⇒ n(n + 10) – 9(n + 10) = 0
⇒ (n + 10)(n – 9) = 0
∴ n + 10 = 0
or n – 9 = 0
∴ n = – 10
or n = 9
∴ n cannot be negative.
∴ n = 9
APPEARS IN
संबंधित प्रश्न
The sum of n, 2n, 3n terms of an A.P. are S1 , S2 , S3 respectively. Prove that S3 = 3(S2 – S1 )
How many multiples of 4 lie between 10 and 250?
Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nth week, her week, her weekly savings become Rs 20.75, find n.
In an AP given an = 4, d = 2, Sn = −14, find n and a.
Find the sum of first 22 terms of an A.P. in which d = 22 and a = 149.
Find the sum of first n odd natural numbers
Find the middle term of the AP 10, 7, 4, ..., (–62).
The 4th term of an AP is 11. The sum of the 5th and 7th terms of this AP is 34. Find its common difference.
Find the sum of the first n natural numbers.
If the ratio of sum of the first m and n terms of an AP is m2 : n2, show that the ratio of its mth and nth terms is (2m − 1) : (2n − 1) ?
Find the A.P. whose fourth term is 9 and the sum of its sixth term and thirteenth term is 40.
The sum of the first n terms of an A.P. is 3n2 + 6n. Find the nth term of this A.P.
If in an A.P. Sn = n2p and Sm = m2p, where Sr denotes the sum of r terms of the A.P., then Sp is equal to
The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is
If the first term of an A.P. is a and nth term is b, then its common difference is
An article can be bought by paying Rs. 28,000 at once or by making 12 monthly installments. If the first installment paid is Rs. 3,000 and every other installment is Rs. 100 less than the previous one, find:
- amount of installments paid in the 9th month.
- total amount paid in the installment scheme.
How many terms of the series 18 + 15 + 12 + ........ when added together will give 45?
Find the sum:
1 + (–2) + (–5) + (–8) + ... + (–236)
If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.
Find the sum of all 11 terms of an A.P. whose 6th term is 30.
