If the pth term of an AP is 1/q and qth term is 1/p find the sun of its first 20 terms
1 answer:
The nth term of an AP = a + (n - 1)d
The pth term of the AP = a +(p - 1)d = 1/q
The qth term of the AP = a +(q - 1)d = 1/p
Sum of nth term = (n/2)(2a + (n - 1)d)
Sum for 20th term = (20/2)(2a + (20- 1)d) = 10(2a + 19d)
a +(p - 1)d = 1/q .....(a)
a +(q - 1)d = 1/p ......(b)
Equation (a) - (b)
a - a + (p - 1)d - (q - 1)d = 1/q - 1/p
pd - d -qd + d = (p - q)/pq
pd - qd = (p - q)/pq
(p - q)d = (p - q)/pq
Cancel out (p - q) from both sides
d = 1/(pq)
Recall a +(p - 1)d = 1/q .....(a)
a +(p - 1)(1/pq)= 1/q
a + (p - 1)/(pq) = 1/q
a = (1/q) - (p - 1)/(pq)
a = (p - (p - 1))/(pq)
a = (p - p + 1) / (pq)
a = 1/(pq)
a = 1/(pq), d = 1/(pq)
Recall sum of 20 terms:
10(2a + 19d)
10(2*(1/(pq)) + 19*(1/(pq)))
10 ( 2/(pq) + 19/(pq)) = 10* (2 + 19)/(pq)
= 10*21/(pq)
= 210 / (pq)
Sum of the 20 terms = 210/(pq)
Hope this explains it.
You might be interested in
N= 8
b/c if u do it backwards 5*1.6=8- you get 8
and to check 8/1.6=5
so n= 5
hope this helps
Answer:
43 1 ÷12
Step-by-step explanation:
The computation of the length of the original roll of papers is shown below:
36 1 ÷3 and 6 3 ÷ 4 together
Now convert the above fractions into a number
36 1 ÷ 3 = 109 ÷3
And,
6 3 ÷4 = 27 ÷ 4
Now add these two numbers i.e.
109 ÷3 + 27 ÷ 4
= 436 + 81 ÷ 12
= 517 ÷12
= 43 1 ÷12
Answer: x-5
Step-by-step explanation: well idk
B, look at picture for explanation
Please let me know if this helped or rate this the brainlist! Thanks