find the AP of first term -4 last term 29 and the sum of its term is 150 find the common difference AP About the author Lyla
Answer Let d be the common difference. Given: first term =a=-4 last term =l=29 Sum of all the terms, [tex] \sf \: S_n=150 \\ \\ \sf \: S_n= \frac{n}{2} [a+l] \\ \\ \sf \implies150= \frac{n}{2} [ – 4+29] \\ \\ \sf \implies \: 150 \times 2 = n \times 25 \\ \\\sf \implies \: 300 = n \times 25 \\ \\ \sf \implies \: n = \cancel \frac{300}{25} \\ \\ \sf \implies \: n=12[/tex] There are 12 terms in total. Therefore, 29 is the 10th term of the AP. Now, [tex] \sf29=a+(12−1)d \\ \\ \sf \implies \: 29= – 4+11d \\ \\ \sf \implies \: 29 + 4=11d \\ \\ \sf \implies \:33 = 11d \\ \\ \sf \implies \: d = \cancel \frac{33}{11} \\ \\ \sf \implies \:d = 3[/tex] The common difference is 3. Reply
Step-by-step explanation:
a = -4
an = 29
S = 150 d = ?
sn=n/2(2a+n-1 d)
150 = 150
Answer
Let d be the common difference.
Given:
Sum of all the terms,
[tex] \sf \: S_n=150 \\ \\ \sf \: S_n= \frac{n}{2} [a+l] \\ \\ \sf \implies150= \frac{n}{2} [ – 4+29] \\ \\ \sf \implies \: 150 \times 2 = n \times 25 \\ \\\sf \implies \: 300 = n \times 25 \\ \\ \sf \implies \: n = \cancel \frac{300}{25} \\ \\ \sf \implies \: n=12[/tex]
There are 12 terms in total.
Therefore, 29 is the 10th term of the AP.
Now,
[tex] \sf29=a+(12−1)d \\ \\ \sf \implies \: 29= – 4+11d \\ \\ \sf \implies \: 29 + 4=11d \\ \\ \sf \implies \:33 = 11d \\ \\ \sf \implies \: d = \cancel \frac{33}{11} \\ \\ \sf \implies \:d = 3[/tex]
The common difference is 3.