Find the number of terms of AP 3,8,13.... so that their sum is 255
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Find the number of terms of AP 3,8,13.... so that their sum is 255
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Answer:
[tex]\bigstar{\bold{Number\:of\:terms=10}}[/tex]
Step-by-step explanation:
[tex]\Large{\underline{\underline{\bf{Given:}}}}[/tex]
[tex]\Large{\underline{\underline{\bf{To\:Find:}}}}[/tex]
[tex]\Large{\underline{\underline{\bf{Solution:}}}}[/tex]
→ We have to find the number of terms of the A.P so that the sum is 255
→ First find the common difference (d) of the A.P
d = a₂ - a₁
d = 8 - 3
d = 5
→ Sum of n terms of a A.P is given by the formula,
[tex]S_n=\dfrac{n}{2}(2a_1+(n-1)\times d)[/tex]
→ Substitute the given datas,
255 = n/2 ( 2 × 3 + (n - 1) ×5)
2 × 255 = n ( 6 + 5n - 5 )
510 = n ( 5n + 1)
5n² + n = 510
5n² + n - 510 = 0
→ Solving it by factorization method
[tex]n=\dfrac{-b\pm \sqrt{b^{2}-4ac } }{2a}[/tex]
where a = 5, b = 1, c = - 510
[tex]n=\dfrac{-1\pm \sqrt{(1)^{2}-4\times 5\times -510 } }{2\times 5}[/tex]
[tex]n=\dfrac{-1\pm\sqrt{10201} }{10}[/tex]
[tex]n=\dfrac{-1\pm 101}{10}[/tex]
→ Case 1 :
[tex]n=\dfrac{-1+101}{10}[/tex]
n = 100/10 = 10
→ Case 2:
[tex]n=\dfrac{-1-101}{10}[/tex]
n = -102/10
→ This is not possible since number of terms cannot be negative or decimals
→ Hence value of n is 10
[tex]\boxed{\bold{Number\:of\:terms=10}}[/tex]
[tex]\Large{\underline{\underline{\bf{Verification:}}}}[/tex]
→ The number of term = 10
S₁₀ = 10/2 ( 2 × 3 + ( 10 - 1) × 5)
S₁₀ = 5 ( 6 + 9 × 5 )
S₁₀ = 5 × 51
S₁₀ = 255
→ Hence proved
[tex]\Large{\underline{\underline{\bf{Notes:}}}}[/tex]
→ The sum of n terms of an A.P is given by the formula
[tex]S_n=\dfrac{n}{2} (2a_1+(n-1)\times d)[/tex]
→ It is also given by the formula,
[tex]S_n=\dfrac{n}{2} (a_1+a_n)[/tex]