find the sum of 1st 10 terms of AP whose 4th term 13 & 8th term is 29​

2 thoughts on “find the sum of 1st 10 terms of AP whose 4th term 13 & 8th term is 29​”

1. 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

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2. 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.

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