find the sum of 1st 10 terms of AP whose 4th term 13 & 8th term is 29
Answers
Answer:
190
Step-by-step explanation:
a4=13
a7=29
a+3d=13
a+7d=29
by solving, d =4 and a=1
S10= 10/2(2×1 +9(4))
= 5(2+36)
=5(38)
= 190
EXPLANATION.
4th term of an A.P. = 13.
8th term of an A.P. = 29.
As we know that,
General term of an A.P.
⇒ Tₙ = a + (n - 1)d.
⇒ T₄ = a + (4 - 1)d.
⇒ T₄ = a + 3d.
⇒ a + 3d = 13. ⇒(1).
⇒ T₈ = a + (8 - 1)d.
⇒ T₈ = a + 7d.
⇒ a + 7d = 29. ⇒(2).
From equation (1) & (2), we get.
⇒ a + 3d = 13.
⇒ a + 7d = 29.
We get,
⇒ -4d = -16.
⇒ 4d = 16.
⇒ d = 4.
Put the value of d = 4 in equation (1), we get.
⇒ a + 3(4) = 13.
⇒ a + 12 = 13.
⇒ a = 13 - 12.
⇒ a = 1.
First term = a = 1.
Common difference = d = b - a = 4.
As we know that,
Sum of nth term of an A.P.
⇒ Sₙ = n/2[2a + (n - 1)d].
⇒ S₁₀ = 10/2[2(1) + (10 - 1)4].
⇒ S₁₀ = 5[2 + 9(4)].
⇒ S₁₀ = 5[2 + 36].
⇒ S₁₀ = 5[38].
⇒ S₁₀ = 190.
MORE INFORMATION.
Supposition of an A.P.
(1) = Three terms as : a - d, a, a + d.
(2) = Four terms as : = a - 3d, a - d, a + d, a + 3d.
(3) = Five terms as : a - 2d, a - d, a, a + d, a + 2d.