How do you find the first three terms of the arithmetic series n=19, #a_n=103#, #S_n=1102#?

3 Answers
May 3, 2018

See explanation.

Explanation:

First we have to write everything which is given and what we are looking for:

Given:

#n=19#

#a_19=103#

#S_19=1102#

To calculate:

#a_1#, #a_2#, #a_3#

First we can use the sum formula to calculate #a_1#:

#S_19=(a_1+a_19)/2*19#

#1102=(a_1+103)/2*19#

#2204=(a_1+103)*19#

#a_1+103=116#

#a_1=13#

Now we can calculate the common difference using two given terms:

#a_1+18*d=a_19#

#13+18*d=103#

#18*d=90#

#d=5#

Now having #a_1# and #d# we can calculate #a_2# and #a_3#:

#a_2=a_1+d=13+5=18#

#a_3=a_2+d=18+5=23#

Answer:

The first three terms are: #13,18# and #23#.

May 3, 2018

#13,18,23.#

Explanation:

Here,

#n=19, a_n=103 , S_n=1102#.

We know that,

#n^(th) term # of Arithmetic series is#=a_n#

and sum of first n-terms#=S_n#

Now, #a=#first term and

#d=#common ratio of Arithmetic series.

So,

#color(red)(a_n=a+(n-1)d and S_n=n/2[2a+(n-1)d]#

#=>103=a+(19-1)d and 1102=19/2[2a+(19-1)d]#

#=>103=a+18d and1102xx2/19=2a+18d#

#=>103=a+18d and 116=2a+18d#

Let,

#color(blue)(a+18d=103 to(1)and 2a+18d=116to(2)#

From #(2)# we get,

#a+color(blue)(a+18d)=116...to[2a=a+a]#

#a+color(blue)(103)=116...tocolor(blue)([use (1) ]#

#=>a=116-103#

#=>color(violet)(a=13#

From #(1) #we have,

#13+18d=103#

#=>18d=103-13#

#=>18d=90#

#=>color(violet)(d=5#

Hence, first three terms of Arithmetic series are :

#(i)a=color(brown)(13#

#(ii)a+d=13+5=color(brown)(18#

#(iii)a+2d=13+2xx5=13+10=color(brown)(23#

May 3, 2018

#" "#
The first three terms: #color(blue)(a_1 = 13, a_2=18 and a_3=23#

Explanation:

#" "#
Total number of terms: #color(red)(n=19#

19th term: #color(red)(a_19=103#

Sum of the first 19 terms: #color(red)(S_19=1102#

In an Arithmetic Sequence, the difference between one term and the next is a Common Difference: #color(red)d#.

The terms are:

#color(brown)(a_1, (a_1+d), (a_1 + 2d), (a_1+3d), ... ,# where #color(brown)(a_1# is the First Term and #color(brown)(d# is the Common Difference.

The Sum of an arithmetic sequence is called an Arithmetic Series.

#color(green)("Step 1:"#

#color(blue)("Formula 1:"#

Sum to #color(red)(n)# terms of an arithmetic series:

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

#color(blue)("Formula 2:"#

Find the #color(red)(n^(th) term# of the arithmetic series:

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

#color(green)("Step 2:"#

Since, #color(red)(n, a_n, and S_n)# are given, use #color(blue)("Formula 1"# to find #color(red)(a_1)#.

#1102=[19(a_1+a_(19))]/2#

#1102=[19(a_1+103)]/2#

#1102=(19a_1+1957)/2#

Multiply both sides of the equation by #color(red)(2#.

#1102*color(red)(2) =(19a_1+1957)/cancel 2*color(red)(cancel 2#

#2204=19a_1+1957#

Flipping sides:

#19a_1+1957=2204#

Subtract #color(red)(1957#.

#19a_1+cancel 1957-color(red)(cancel 1957)=2204-color(red)(1957)#

#19a_1=247#

Divide both sides by #color(red)(19#

#(cancel 19a_1)/color(red)(cancel 19)=247/color(red)(19#

#a_1=247/19=13#

# :. "First Term "= a_1 = 13#

#color(green)("Step 3:"#

Use #color(blue)("Formula 2"# to find the Common Difference #color(red)(d)#.

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

When #n=19, a_n = a_19#

#103=13+d(19-1)#

#103=13+19d-d#

#103=13+18d#

Flipping sides:

#13+18d=103#

Subtract #color(red)13#.

#cancel 13+18d-color(red)cancel 13=103-color(red)13#

#18d=90#

Divide by #color(red)(18#.

#(cancel 18d)/color(red)(cancel 18)=cancel 90^color(green)(5)/color(red)(cancel 18#

# :. "Common Difference " color(red)(d=5#

Terms: #13+[13+1(5)]+[13+2(5)]#

So, #a_1 = 13, a_2=18 and a_3=23#