An AP consits of 37 terms …

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Sia ? 6 years, 4 months ago
Let the middle most terms of the A.P. be {tex}(a - d), a, (a + d){/tex}
Given {tex}a -d + a + a + d = 225{/tex}
{tex}3a = 225{/tex}
or, {tex}a = 75{/tex}
and the middle term = {tex}\frac { 37 + 1 } { 2 }{/tex}= 19th term
{tex}\therefore{/tex} A.P. is
(a - 18d ),....(a - 2d), {a - d), a, (a + d), (a + 2d),..........{tex}(a + 18d){/tex}
Sum of last three terms
{tex}(a + 18d) + (a + 17d) + (a + 16d) = 429{/tex}
or,{tex} 3a +51d = 429{/tex}
or, {tex}225 + 51d = 429 {/tex}
or, d = 4
First term a1 = a - 18d = 75 - 18{tex}\times{/tex}4 = 3
a2 = 3 + 4 = 7
Hence, A.P. = 3, 7, 11 ,.........., 147
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