WebApr 15, 2024 · Best answer Let d be the common difference. Given: first term = a = -4 last term = l = 29 Sum of all the terms = Sn = 150 Sn = n/2 [a + l] 150 = n/2 [-4 + 29] n = 12 There are 12 terms in total. Therefore, 29 is the 12th term of the AP. Now, 29 = -4 + (12 – 1)d 29 = -4 + 11d d = 3 The common difference is 3. ← Prev Question Next Question → WebSep 2, 2024 · Best answer i. Sum of first five terms = 150 Sum of the five consecutive terms of arithmetic sequence is five times of its middle term. Third term = 150 5 = 30 150 5 = 30 ii. First term + Tenth term = Second term + Nineth term = Third term + Eighth term = Fourth term + Seventh term = Fifth term + Sixth term = 550 5 = 110 550 5 = 110
In an AP of 50 terms the sum of first 10 terms is 210 and the sum …
WebThe sum of first 10 terms of an AP is -150 and the sum of its next 10 terms is -550. Find the AP. Solution Let a be the first term and d be the common difference of the AP. Then, It is … WebAnswer (1 of 5): 15+30+45 … +150 is an AP, whose first term, a = 15 and the common difference = 15. Tn = 150 = a+(n-1)d = 15 + (n-1)*15, or dividing right through by 15, 10 = 1 + (n-1), or n = 10 Sn = (n/2)[2a + (n-1)d] S10= (10/2)[2*15 + … duty of care teachers
In an A.P., the Sum of First Ten Terms is −150 and the Sum of Its …
WebJul 29, 2024 · In an AP, the sum of first ten terms is -150 and the sum of its next ten terms is -550. Find the A.P.I have provided the easiest solution of above question, ... WebOpen in App Solution Verified by Toppr From the question it is given that, The sum of the first 10 terms =−150 The sum of the next 10 terms =−550 A.P =? We know that, Sn … WebYou would do the exact same process, but you would have to SOLVE for "n" (number of terms) first. To do so, you must start with the arithmetic sequence formula: tn = a + d(n −1) Then, sub in all known values. tn = 15 (last term of the sequence), a = 1 (first term), d = 2 (difference between terms) and solve for n like so: 15 = 1 + 2(n −1) duty of care to patients