D=3, a10=27.5 Sn=? a1= ?, n=?

Arithmetic Sequence , #a_n = a + (n-1) d#

#a_n # is #n^(th)# term, #a# the first term and #d# the common difference.

Given #a_n = 27.5, d = 3#, to find sum of first n terms #S_n#

#S_n = n/2 (2a + (n-1) d#

#a + (10-1)*3 = 27.5#

#a = 27.5 - (10-1)*3 = 27.5 - 27 = 0.5#

Sum of first 10 terms, #S_10 = (10/2) * (2*0.5 + (10-1)*3)#

#S_10 = 5 * (1 + 9*3) = 140#

Verification:

#S_n = n/2 (a + l)# where l is the last term.

#S_10 = (10/2) * (a + a_10)#

#S_10 = 5 * (0.5 + 27.5) = 140#

#" "#
Given :

Common difference : #color(red)(d=3#

Tenth term of the arithmetic sequence: #color(red)(a_10=27.5#

#color(red)("What is expected ?"#

Sum to "n" terms : #color(blue)(S_n#

First term : #color(blue)(a_1#

Total number of terms in the arithmetic series : #color(blue)(n#

Using #color(red)(d and a_10#, we can find the first term, which is indicated by #color(red)(a_1#

Formula used to find a specific term:

#color(blue)(a_n=a_1+(n-1)d#

We will use this formula to find the first term : #color(red)(a_1#

Substitute all known values in the formula:

#color(blue)(27.5=a_1+(10-1)3#

#rArr 27.5=a_1+(9)(3)#

#rArr 27.5=a_1+27#

Subtract #color(blue)(27# from both sides of the equation to isolate #color(blue)(a_1#

#rArr 27.5-27=a_1+27-27#

#rArr 0.5=a_1+cancel 27-cancel 27#

#rArr color(red)(a_1=0.5#

Hence, we have our first term of the arithmetic series: #color(red)(a_1=0.5#

We now have the following useful details:

First term : #color(red)(a_1=0.5#

Tenth term: #color(red)(a_10=27.5#

Common difference : #color(red)(d=3#

The formula to find the Sum of "n" terms in any arithmetic series is given by:

#color(red)(S_n=[n(a_1+a_n)]/2#

OR

#color(red)(S_n=[(n/2)(a_1+a_n)]#

We can find the SUM to 20, 50, 100 or more terms of the arithmetic series using the above formula:

Assume that we want to find the SUM to 100 terms:

i.e., find #color(red)(S_100#

We will use the following formula to find a specific term:

#color(blue)(a_n=a_1+(n-1)d#

We know that #a_1=0.5; d = 3; n=100#

We can find #color(red)(a_100#

#rArr a_100 =a_1+(n-1)3#

#rArr a_100=0.5+(100-1)3#

#rArr a_100=0.5+(99)(3)#

#rArr a_100=0.5+297#

#rArr a_100=297.50#

To find the sum to 100 terms use the formula:

#color(red)(S_n=[(n/2)(a_1+a_n)]#

#rArr S_100=(100/2)(0.5+297.50)#

#rArr S_100=(50)(298)#

Hence, #color(red)(S_100 = 14,900#

Hope you find this solution useful.