If sum of 'n' terms of an A.P . is 3n2+5n and am=164 , then m=
Answers
m = 27
Step-by-step explanation:
Sn = 3n2+5n
putting value for 'n' as 1,2,3..
S1 = 3*1'2 + 5*1
= 3+5
= 8 = a
S2 = 3*2'2 + 5*2
= 3*4 + 10
= 12 + 10
= 22
t2 = 22-8
= 14
d = 14-8
= 6
am = 164
164 = a + (m-1)d
164 = 8 + (m-1)6
164-8/6 = (m-1)
26 = (m-1)
26+1 = m
27 = m
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EXPLANATION.
Sum of n terms of an A.P. = 3n² + 5n.
As we know that,
⇒ Tₙ = Sₙ - Sₙ₋₁.
⇒ 3n² + 5n - [3(n - 1)² + 5(n - 1)].
⇒ 3n² + 5n - [3(n² + 1 - 2n) + 5n - 5].
⇒ 3n² + 5n - [3n² + 3 - 6n + 5n - 5].
⇒ 3n² + 5n - [3n² - n - 2].
⇒ 3n² + 5n - 3n² + n + 2.
⇒ 5n + n + 2.
⇒ 6n + 2 = Algebraic expression.
As we know that,
Put the value of n = 1 in equation, we get.
⇒ 6(1) + 2.
⇒ 6 + 2.
⇒ 8.
Put the value of n = 2 in equation, we get.
⇒ 6(2) + 2.
⇒ 12 + 2.
⇒ 14.
Put the value of n = 3 in equation, we get.
⇒ 6(3) + 2.
⇒ 18 + 2.
⇒ 20.
Put the value of n = 4 in equation, we get.
⇒ 6(4) + 2.
⇒ 24 + 2.
⇒ 26.
Their Series = 8, 14, 20, 26,,,,,,
First term of an A.P. = a = 8.
Common difference of an A.P. = d = b - a = 14 - 8 = 6.
As we know that,
General equation of an A.P.
⇒ Tₙ = a + (n - 1)d.
⇒ T(m) = a + (m - 1)d.
⇒ 164 = a + (m - 1)d.
⇒ 164 = 8 + (m - 1)6.
⇒ 164 = 8 + 6m - 6.
⇒ 164 = 6m + 2.
⇒ 164 - 2 = 6m.
⇒ 162 = 6m.
⇒ m = 27.
MORE INFORMATION.
Supposition of terms in 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.