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the fourth term of an AP is 5 while the sum of the first 6 terms is 10. Find the sum of the first 19 terms​

User Kennydust
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1 Answer

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Answer: S₁₉ = 855

Explanation:

T₄ = a + ( n - 1 )d = 5 , from the statement above , but n = 4

a + 3d = 5 -------------------------1

S₆ = ⁿ/₂[(2a + ( n - 1 )d] = 10, where n = 6

= ⁶/₂( 2a + 5d ) = 10

= 3( 2a + 5d ) = 10

= 6a + 15d = 10 -----------------2

Now solve the two equation together simultaneously to get the values of a and d

a + 3d = 5

6a + 15d = 10

from 1,

a = 5 - 3d -------------------------------3

Now put (3) in equation 2 and open the brackets

6( 5 - 3d ) + 15d = 10

30 - 18d + 15d = 10

30 - 3d = 10

3d = 30 - 10

3d = 20

d = ²⁰/₃.

Now substitute for d to get a in equation 3

a = 5 - 3( ²⁰/₃)

a = 5 - 3 ₓ ²⁰/₃

= 5 - 20

a = -15.

Now to find the sum of the first 19 terms,

we use the formula

S₁₉ = ⁿ/₂( 2a + ( n - 1 )d )

= ¹⁹/₂( 2 x -15 + 18 x ²⁰/₃ )

= ¹⁹/₂( -30 + 6 x 20 )

= ¹⁹/₂( -30 + 120 )

= ¹⁹/₂( 90 )

= ¹⁹/₂ x 90

= 19 x 45

= 855

Therefore,

S₁₉ = 855

User Pattabi Raman
by
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