WebMar 12, 2024 · Sum of n terms in an arithmetic progression is given by the formula S = n 2 [ 2 a + ( n − 1) d] in which a = first term, n = number of terms and d = common difference. Let us understand this concept in brief by taking an example. Consider a man putting 100 rupees in his daughter’s piggy bank, in such a way that, He deposits 100 rupees on ... WebExpert Answers. hala718. Certified Educator. Share Cite. Le a1, a2, ..., a10 are terms of an A.P. given S10 = a1+a2+...+10 = 150. But we know that: s10 = (a1+a10)*10/2 = 150. ==> …
Sums of Arithmetic Sequences - Precalculus Socratic
WebMar 27, 2024 · C. I Frequency 10 vicle, me ACB be 0-15 15-30 30-45 45-60 ore than type ogive' for the given data. 18 40 20 60-75 12 29) The sum of the reciprocals of Rehman's ages 3years ago and 5 years from now is Find his present age. WebSolution 1 Here , a = 2 , l= 29 and s n = 155 Let d be the common difference of the given AP and n be the total number of terms. Then, T n = 29 ⇒ a + (n-1) d = 29 ⇒ 2 + (n-1) d= 29 .............. (1) The sum of n terms of an AP is given by s n = n 2 [ a + l] = 155 ⇒ n 2 [ 2 + 29] = ( n 2) × 31 = 155 ⇒ n = 10 Putting the value of n in (i), we get: itinct.com
In an A.P., the sum of first ten terms is −150 and the sum of its …
Webasked Sep 14, 2024 in Mathematics by Mubarak (32.8k points) In an AP, the first term is -4, the last term is 29 and the sum of all its terms is 150. Find its common difference. arithmetic progression class-10 1 Answer +1 vote answered Sep 14, 2024 by AmirMustafa (60.3k points) selected Sep 23, 2024 by Vikash Kumar Best answer WebOct 20, 2024 · Let a be the first term and d be the common difference of the given AP . S₁₀ = -150. ⇒ Sn = n/2 [ 2a + (n-1)d] ⇒ S₁₀ = 10/2 [ 2a + ( 10 - 1 ) d ]. ⇒ -150= 10/2 [ 2a + 9d ] ⇒ … WebApr 11, 2024 · Given the sum of the first 10 terms of an AP is -150 and the sum of the next ten terms which means Sum of first 20 terms-Sum of first 10 terms is -530.. Let the first term, common difference of the AP be a and d respectively. Now, ⇒ S₁₀ = 10/2 [ 2a + (10 - 1)d ] ⇒ -150 = 5 [ 2a + 9d ] ⇒ 2a + 9d =-30...(i)Now, according to the second case, itincta