sum of first and fifth term of an ap is 18 and 4th term exists twice first term by 2 find the sum of first 20 terms
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given,,
a+a+4d = 18
2a+4d = 18
a+2d = 9 (i)
given,,
a+3d =2a+2
3d - a = 2 (ii)
3d = 2 + a
d = 2 + a/3
Put value of d in (i)
a+2 (6+a/3) = 9
a + 12+2a/3 = 9
3a + 12 + 2a/ 3 = 9
5a + 12 = 27
5a = 15
a = 3
d = 5/3
Sum = n/2[2a+ (n-1) d]
10 [ 6+ 19× 5/3 ]
10 [18+95/3]
10× 113/3
1130/3 = 377 ( approx...)
a+a+4d = 18
2a+4d = 18
a+2d = 9 (i)
given,,
a+3d =2a+2
3d - a = 2 (ii)
3d = 2 + a
d = 2 + a/3
Put value of d in (i)
a+2 (6+a/3) = 9
a + 12+2a/3 = 9
3a + 12 + 2a/ 3 = 9
5a + 12 = 27
5a = 15
a = 3
d = 5/3
Sum = n/2[2a+ (n-1) d]
10 [ 6+ 19× 5/3 ]
10 [18+95/3]
10× 113/3
1130/3 = 377 ( approx...)
Answered by
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Answer:
a = 23/5 ...d = 11/5
Step-by-step explanation:
take the value and substitute correctly the previous answer was wrong..who did that they didn't took the values correctly
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