An A.P consists of 50terms of …
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Sia ? 6 years, 6 months ago
Let the first term and the common difference of the AP be a and d respectively.
3rd term = 12 ..... Given
{tex} \Rightarrow {/tex} a + (3 - 1)d = 12 {tex}\because {a_n} = a + (n - 1)d{/tex}
{tex} \Rightarrow {/tex} a + 2d = 12 ......... (1)
Last term = 106 ........ Given
{tex} \Rightarrow {/tex} So the term = 106 {tex}\because {/tex} The AP consists of 50 terms
{tex} \Rightarrow {/tex} a + (50 - 1)d = 106
{tex} \Rightarrow {/tex} a + 49d = 106 ........ (2)
Solving (1) and (2), we get
a = 8
d = 2
Therefore, 29th term of the AP
= 9 + (29 - 1)d {tex}\because {a_n} = a + (n - 1)d{/tex}
= 9 + 28d
= 8 + (28) (2)
= 8 + 56
= 64
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