If third term and fifth term of an A.P. are 13 and 25 respectively, find its 7th term. (A) 30 (B) 33 (C) 37 (D) 3 About the author Remi
Answer: 37 Step-by-step explanation: if there is difference of 12 in 3rd and 5th term then difference between 5th and 7th term is 12… as it is in ap so answer will be 37 I hope it help u, plz mark this answer as brainliest answer Reply
Solution Given :– If third term and fifth term of an A.P. are 13 and 25 Find :– 7th terms of A.P. Explanation Let, First terms = a Common Defference = d Using Formula [tex]\boxed{\underline{\tt{\red{\:T_{n}\:=\:a+(n-1)d}}}}[/tex] Where n = Number of terms Case 1. If, n = 3 ==> T3 = a + (3 – 1)d ==> 13 = a + 2d___________(1) Case 2. If, n = 5 ==> T5 = a + (5 – 1)d ==> 25 = a + 4d_________(2) Subtract equ(1) & equ(2) ==> -2d = -12 ==> d = 12/2 ==> d = 6. keep in equ(2) ==> a + 4 × 6 = 25 ==> a = 25 – 24 ==> a = 1 So,Now calculate 7th terms Where a = 1 d = 6 ==> T7 = a + 6d keep Value of a & d ==> T7 = 1 + 6 × 6 ==> T7 = 1 + 36 ==> T7 = 37 Hence 7th terms will be = 37 ____________________ Reply
Answer:
37
Step-by-step explanation:
if there is difference of 12 in 3rd and 5th term then difference between 5th and 7th term is 12…
as it is in ap
so answer will be 37
I hope it help u, plz mark this answer as brainliest answer
Solution
Given :–
Find :–
Explanation
Let,
Using Formula
[tex]\boxed{\underline{\tt{\red{\:T_{n}\:=\:a+(n-1)d}}}}[/tex]
Where
Case 1.
==> T3 = a + (3 – 1)d
==> 13 = a + 2d___________(1)
Case 2.
==> T5 = a + (5 – 1)d
==> 25 = a + 4d_________(2)
Subtract equ(1) & equ(2)
==> -2d = -12
==> d = 12/2
==> d = 6.
keep in equ(2)
==> a + 4 × 6 = 25
==> a = 25 – 24
==> a = 1
So,Now calculate 7th terms
Where
==> T7 = a + 6d
keep Value of a & d
==> T7 = 1 + 6 × 6
==> T7 = 1 + 36
==> T7 = 37
Hence
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