if 8th term of an AP is equal to 5 terms its second term .if the 5th term is 8 find the first 10 terms
term8 = a+7d
term2 = a+d
a+7d = 5(a+d)
a + 7d = 5a + 5d
2d = 4a
d = 2a
term5 = a+4d = 8
a + 4(2a) = 8
9a = 8
a = 8/9 ---> d = 16/9
terms are:
8/9, 24/9, 40/9, 56/9, 72/9, 88,9, 104/9, 120/9, ...
carry on
check:
5 times the 2nd
= 5(24/9) = 120/9 which is term 8
To find the first 10 terms of an arithmetic progression (AP), we first need to determine the common difference (d) of the AP.
Given that the 8th term of the AP is equal to 5 times its second term, we can set up the equation:
a + 7d = 5(a + d)
Here, 'a' represents the first term.
Expanding and simplifying the equation:
a + 7d = 5a + 5d
7d - 5d = 5a - a
2d = 4a
d = 2a
We are also given that the 5th term of the AP is 8, which means:
a + 4d = 8
Substituting the value of d in terms of a into this equation:
a + 4(2a) = 8
a + 8a = 8
9a = 8
a = 8/9
Now, we have found the value of the first term (a) as 8/9.
To find the first 10 terms, we can substitute the value of a into the expression for the nth term of an arithmetic progression:
tn = a + (n-1)d
Substituting n = 1 to 10 into the expression, with a = 8/9 and d = 2a:
t1 = 8/9 + (1-1)(2(8/9)) = 8/9
t2 = 8/9 + (2-1)(2(8/9)) = 16/9
t3 = 8/9 + (3-1)(2(8/9)) = 24/9 = 8/3
t4 = 8/9 + (4-1)(2(8/9)) = 32/9
t5 = 8/9 + (5-1)(2(8/9)) = 40/9
t6 = 8/9 + (6-1)(2(8/9)) = 48/9 = 16/3
t7 = 8/9 + (7-1)(2(8/9)) = 56/9
t8 = 8/9 + (8-1)(2(8/9)) = 64/9
t9 = 8/9 + (9-1)(2(8/9)) = 72/9 = 8
t10 = 8/9 + (10-1)(2(8/9)) = 80/9
Therefore, the first 10 terms of the AP are:
8/9, 16/9, 8/3, 32/9, 40/9, 16/3, 56/9, 64/9, 8, 80/9