the 5th and 12th term of an AP is 23cand 37 find the sum of 40 terms
Answers
Answer:
- Answer:a+4d=23
- Answer:a+4d=23a+11d=57
Answer:a+4d=23a+11d=57 then,a=23-4d
Answer:a+4d=23a+11d=57 then,a=23-4dput the value of a in eq2
- Answer:a+4d=23a+11d=57 then,a=23-4dput the value of a in eq2a+11d=57
Answer:a+4d=23a+11d=57 then,a=23-4dput the value of a in eq2a+11d=57 23-4d+11d=57. (a=23-4d)
Answer:a+4d=23a+11d=57 then,a=23-4dput the value of a in eq2a+11d=57 23-4d+11d=57. (a=23-4d) 7d=34, d=34/7
Answer:a+4d=23a+11d=57 then,a=23-4dput the value of a in eq2a+11d=57 23-4d+11d=57. (a=23-4d) 7d=34, d=34/7 a=23-34/7
Answer:a+4d=23a+11d=57 then,a=23-4dput the value of a in eq2a+11d=57 23-4d+11d=57. (a=23-4d) 7d=34, d=34/7 a=23-34/7 a=127/7, d=34/7
Answer:a+4d=23a+11d=57 then,a=23-4dput the value of a in eq2a+11d=57 23-4d+11d=57. (a=23-4d) 7d=34, d=34/7 a=23-34/7 a=127/7, d=34/7 sum of 40 terms
Answer:a+4d=23a+11d=57 then,a=23-4dput the value of a in eq2a+11d=57 23-4d+11d=57. (a=23-4d) 7d=34, d=34/7 a=23-34/7 a=127/7, d=34/7 sum of 40 terms S44=n/2{(2a +(n-1)d)}
Answer:a+4d=23a+11d=57 then,a=23-4dput the value of a in eq2a+11d=57 23-4d+11d=57. (a=23-4d) 7d=34, d=34/7 a=23-34/7 a=127/7, d=34/7 sum of 40 terms S44=n/2{(2a +(n-1)d)} S44=40/2{2×127/7)+(40-1)34/7}
Answer:a+4d=23a+11d=57 then,a=23-4dput the value of a in eq2a+11d=57 23-4d+11d=57. (a=23-4d) 7d=34, d=34/7 a=23-34/7 a=127/7, d=34/7 sum of 40 terms S44=n/2{(2a +(n-1)d)} S44=40/2{2×127/7)+(40-1)34/7} S44=20{254/7+1326/7}
Answer:a+4d=23a+11d=57 then,a=23-4dput the value of a in eq2a+11d=57 23-4d+11d=57. (a=23-4d) 7d=34, d=34/7 a=23-34/7 a=127/7, d=34/7 sum of 40 terms S44=n/2{(2a +(n-1)d)} S44=40/2{2×127/7)+(40-1)34/7} S44=20{254/7+1326/7} S44=20{1580/7}
Answer:a+4d=23a+11d=57 then,a=23-4dput the value of a in eq2a+11d=57 23-4d+11d=57. (a=23-4d) 7d=34, d=34/7 a=23-34/7 a=127/7, d=34/7 sum of 40 terms S44=n/2{(2a +(n-1)d)} S44=40/2{2×127/7)+(40-1)34/7} S44=20{254/7+1326/7} S44=20{1580/7} S44=20×1580/7
S44=31600/7
S44=4514.28571 or4514.29