An AP 5, 8, 11…has 40 terms. Find the last term. Also find the sum of the last 10 terms.
Answered Feb 05, 2023
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
Correct Answer - `(n)/(6)(4n^(2)-3n+23)`
Correct Answer - (i) 85 (ii) -82 (iii) `35sqrt(2) (iv) 4(3)/(4)` (v) 310
Correct Answer - (i) 893 (ii) -56 (iii) - 80 (iv) `(33)/(20)` (v) 5505
Correct Answer - `(i) T_(n) = (5n-1), T_(20) = 99 (ii) T_(n) = (3n +1), T_(25) = 76`
Pls write full question.Your question is a bit ambiguous.
Correct Answer - 1,-1,2
Let the numbers be ‘a’ and ‘b’. Given, sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. ⇒ a + b = 8 ⇒ a...
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