Advertisements
Advertisements
प्रश्न
Find the sum of first 25 terms of the A.P. whose nth term is given by an = 5 + 6n. Also, find the ratio of 20th term to 45th term.
Advertisements
उत्तर
Given that,
`\implies` an = 5 + 6n
We have,
`\implies` a1 = 5 + 6(1) = 11
`\implies` a2 = 5 + 6(2) = 17
So, a = 11, d = 6
Sum of first 25 terms = `n/2(2a + (n - 1)d)`
= `25/2[2(11) + (25 - 1)6]`
= `25/2[22 + 144]`
= `25/2[166]`
= 2075
Now, a20 = a + 19d
= 11 + 19(6)
= 125
`\implies` a45 = a + 19d
= 11 + 44(6)
= 275
Required ratio = `a_20/a_45`
= `125/275`
= `5/11`
Ratio is 5:11.
संबंधित प्रश्न
The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.
If the sum of first n terms is (3n2 + 5n), find its common difference.
Find the sum of all 2 - digit natural numbers divisible by 4.
Which term of the sequence 114, 109, 104, ... is the first negative term?
Two A.P.'s have the same common difference. The first term of one of these is 8 and that of the other is 3. The difference between their 30th term is
Q.19
Find the sum of first 1000 positive integers.
Activity :- Let 1 + 2 + 3 + ........ + 1000
Using formula for the sum of first n terms of an A.P.,
Sn = `square`
S1000 = `square/2 (1 + 1000)`
= 500 × 1001
= `square`
Therefore, Sum of the first 1000 positive integer is `square`
Find t21, if S41 = 4510 in an A.P.
The first term of an AP is –5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.
Find the sum:
1 + (–2) + (–5) + (–8) + ... + (–236)
