8. First term of an arithmetic progression is 8, nth
term is 33 and sum of first n terms is 123, then
find n and common difference d.
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8. First term of an arithmetic progression is 8, nth
term is 33 and sum of first n terms is 123, then
find n and common difference d.
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[tex]\huge\mathfrak{Answer:}[/tex]
Given:
To Find:
Solution:
Given,
[tex] \sf{a = 8}[/tex]
[tex] \sf{an = 33}[/tex]
[tex] \sf{sn = 123}[/tex]
We know that,
[tex] \sf{an = a +(n - 1) \times d}[/tex]
Substituting the values, we have
[tex] \implies\sf{ 33 = 8 + (n - 1)d}[/tex]
[tex]\implies\sf{ 33 - 8 = (n - 1)d}[/tex]
[tex]\implies\sf{ 25 = (n - 1)d}[/tex]
Now, we have
[tex] \sf{sn = \dfrac{n}{2} [{2a + (n - 1)d}]}[/tex]
Substituting the values, we have
[tex] \sf{123 = \dfrac{n}{2} [2 \times 8 + 25]} [/tex]
[tex]\implies\sf{ 123 \times 2 = n(2 \times 8 + 25)}[/tex]
[tex]\implies\sf{ 246 =n( 16 + 25)}[/tex]
[tex]\implies\sf{246 = n(41)}[/tex]
[tex]\implies\sf{ \dfrac{246}{41} = n}[/tex]
[tex]\implies\sf{ n = 6}[/tex]
Now, we can find the value of d by substituting the value of n in n(n - 1)d = 25.
[tex]\implies\sf{ (6 - 1)d = 25}[/tex]
[tex]\implies\sf{ (5)d = 25}[/tex]
[tex]\implies\sf{ d = \dfrac{25}{5}} [/tex]
[tex]\implies\sf{ d = 5}[/tex]
.Hence, the value of n is 6 and the value of d is 5.
Answer:
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