Question

The sum of first 14 terms of 1,1.4,1.8,.... is equal to the sum of first n terms of 5.6,5.8,6.0,.... , find the value of n

Answer 1

these both are arithmetic progressions cuz they both have common differences.
form the first sequence 1, 1.4, 1.8 ,..... we have a common difference of 0.4
snow the sum of 14 terms is
n/2[2(a1) + (n-1)d]
here n =14
when you will find this take the second sequence
5.6,5.8,6.0,....
find sum of n terms by
n/2[2(a1) + (n-1)d]
here n = unknown and d = 0.2
now equate these both sums and find n

Answer 2

just use the formula to sum up to nth term of an Arithmetic sequence
which is
S=(n/2)(2*firsterm+(n-1)*d)

Answer 3

as per the question
we have
(14/2)*(2*1+(14-1)*0.6)=(n/2)(2*5.6 +(n-1)*.2)

Answer 4

i cant simplify further help plz

Answer 5

hello anyone there?

Answer 6

\[50.4=\frac{ n }{ 2 }[(2*5.6)+{(n-1)*0.2}]\]

Answer 7

i am having problem solving this

Answer 8

a partial sum is calculated this way:
\[S_{n} = \frac{ n(a_{1} + a_{n}) }{ 2 }\]

in the first series you were given 14 terms, a first term, but not the last term. we need all three of these values to solve for the partial sum.

to solve for the last term in the first series use the fact that

\[a_{n} = a_{1} + (n-1)d\]

thus an = 1 + (14 - 1)*0.4 = 6.2

now you can calculate the sum of the first series

Sn = 14(1+ 6.2)/2 = 50.4

this sum is the same as the sum of the second series

using the two equations again

an = a1 + (n-1) * d
an = 5.6 + (n - 1) * 0.2

and

Sn = n(a1 + an)/2
Sn = n(5.6 + an)/2
50.4 = n(5.6 + an)/2

we have a system of two equations with two variables

an = 5.6 + (n - 1) * 0.2
50.4 = n(5.6 + an)/2

solve the system to determine n