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An AP 5, 8, 11…has 40 terms. Find the last term. Also find the sum of the last 10 terms. |
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Answer» First Term of the AP (a) = 5 Common difference (d) = 8 - 5 = 3 Last term = a40 = a + (40 - 1) d = 5 + 39 × 3 = 122 Also a31 = a + 30d = 5 + 30 × 3 = 95 Sum of last 10 terms = \(\frac{n}{2}\)(a31 + a40) = \(\frac{10}{2}\)(95 + 122) = 5 × 217 = 1085 |
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