2-5) The sum of terms of no arithmetic progressions are in the ratio (2n+11) : (5n+14) Find the ratio of their pth terms.
pls answer correctly fast
Answers
Answer:
We know that the sum of n terms of an arithmetic progression (AP) is given by:
2
n
×(2a+(n−1)d)
where a= first term of the AP
d= common difference of the AP
Let,
a
1
,d
1
= first term and common difference of the first AP
b
1
,d
2
= first term and common difference of the second AP
Hence,
(2b
1
+(n−1)d
2
)
(2a
1
+(n−1)d
1
)
=
(7n+15)
(3n+8)
---(1)
We need to find out ratios of 12
th
terms of both APs
which is
b
1
+11d
2
a
1
+11d
1
--(2).
Multiply and divide equation (2) with 2
which is
2b
1
+22d
2
2a
1
+22d
1
--(3)
On comparing coefficients of common differences we get n−1=22
⇒n=23.
I.e., if we put n=23 in equ (1) to get ratios of 12
th
terms.
i) Ratios of 12
th
terms =
(7(23)+15)
(3(23)+8)
=
176
77
=
11
7
Ratios of 15
th
terms =
b
1
+14d
2
a
1
+14d
1
--(3).
Divide and multiply 2 with equation 3, which is
2b
1
+28d
2
2a
1
+28d
1
--(4)
On comparing coefficients of common differences we get n−1=28
On substituting n=29 in equation (1). We get
iI) Ratios of 15
th
terms =
(7(29)+15)
(3(29)+8)
=
218/95 answer
Answer:
-5) The sum of terms of no arithmetic progressions are in the ratio (2n+11) : (5n+14) Find the ratio of their pth terms.
mark itsplover Brainlist
he has answer correct