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प्रश्न
Find the 12th from the end of the A.P. – 2, – 4, – 6, …; – 100.
योग
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उत्तर
A.P. = – 2, – 4, – 6, …; – 100
a = –2, d = –4 – (–2)
= –4 + 2
= –2
l = –100
∴ Tn = a + (n – 1)d
⇒ –100 = –2 + (n – 1) x (–2)
⇒ –100 = –2 – 2n + 2
⇒ +2n = 100
⇒ n = `(100)/(2)` = 50
Let mth term is the 12th term from the end
Then m = l – (n – 1)d
= – 100 – (12 – 1) x (– 2)
= – 100 + 22
= – 78.
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