find the number of terms in an A.p given that it's 1st and last terms are a and 37a respectively, and that it's common difference is 49
last term is a + (n-1)d = 37a
36a = 49(n-1)
36a = 49n - 49
n = (36a+49)/49 or (36/49)a + 1
but n must be an integer, so a must be a multiple of 49
if a=49 , n = 37
if a = 98 , n = 73
if a = 147 , n = 109
etc
(n increases by 36)
check my second example, a = 98 , n = 73
terms would be 98 , 147, 196 , ...
term(71) = a + 72d = 98 + 72(49) = 3626
and 37a = 37(98) = 3626
If the sum of 8th and 9th terms of an arithematic prograssion is 72 and the 4th term is -6, find the common difference.
I need the solution pls
To find the number of terms in an arithmetic progression (A.P.), we can use the formula:
n = (last term - first term) / common difference + 1
In this case, the first term (a) is given as a, the last term (37a) is given as 37a, and the common difference is given as 49.
Using the formula, we can substitute the values into the formula and calculate the number of terms:
n = (37a - a) / 49 + 1
n = 36a / 49 + 1
n = (36a + 49) / 49
So, the number of terms in the arithmetic progression is (36a + 49) / 49.