Find the 30th term of A.p …
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Sia ? 6 years, 6 months ago
We have
a = 10, d = (7-10) = -3, l= -62 and n = 11.
{tex}\therefore{/tex}11th term from the end = [l - (n -1) {tex}\times{/tex} d]
= {-62 - (11 -1) {tex}\times{/tex} (-3)}
= (-62 + 30) = -32.
Hence, the 11th term from the end of the given AP is -32.
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