WebApr 2, 2024 · The median of the distribution given below is 14.4. Find the values of x and y, if the sum of frequency is 20. Find the common difference ' d ' of an AP whose first term is 10 and the sum of the first 14 terms is 1505 . For what value of ' n ', are the nth terms of the APs : 9,7,5,….. and 15,12,9,…. the same? WebIn an Ap of 50 terms the sum of first 10 terms is 210 and sum of its 15 terms is 2565.find the ap Report ; Posted by Amena Haseen 5 years, 5 months ago. CBSE > Class 10 > Mathematics ... Xi 0-25 25-50 50-75 75-100 100-125 125-150 150- 175 xi 4 8 x 15 10 7 5 and median is 90 find x. Report ; Posted by Manas Dhiman 1 day, 18 hours ago.
Find the common difference
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 WebOct 20, 2024 · In an AP the sum of first ten terms is -150 and the sum of its next 10 terms is -550. Find the AP. Is this the question you’re looking for? Advertisement Expert-Verified Answer 49 people found it helpful Hey there !! 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] the outlet cheam
In an AP of 50 terms, the sum of the first 10 terms is 210
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 + … WebSum In an A.P., the sum of first ten terms is −150 and the sum of its next ten terms is −550. Find the A.P. Advertisement Remove all ads Solution Here, we are given S 10 = -150 and sum of the next ten terms is −550. Let us take the first term of the A.P. as a and the common difference as d. So, let us first find a10. WebApr 8, 2024 · Given sum of first ten terms = − 150. We know that Sum of n terms of AP S n = n 2 [ 2 a + ( n − 1) d] Therefore, ⇒ S 10 = 10 2 [ 2 a + ( 10 − 1) d] ⇒ − 150 = 5 [ 2 a + 9 d] ⇒ … the outlet challenge