the sum of first three terms of an arithmetic series is 42 and that of the first terms is 80 find the 20th term of series
Answers
Step-by-step explanation:
*Let the first three terms of the A.P be a - d, a , a + d.
We are given that the sum of first three terms of the A.P is 42.
\longmapsto \rm{a \cancel{ - d} + a + a + \cancel d = 42}⟼a−d+a+a+d=42
\longmapsto \rm{3a = 42}⟼3a=42
\longmapsto \rm{a = \dfrac{ \cancel{42}}{ \cancel3} }⟼a=342
\longmapsto \rm{ \pink{a = 14} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1)}⟼a=14(1)
According to the second statement we know that sum of the first five terms of the A.P is 80.
Now, let the A.P be a ,a + d, a + 2d, a + 3d, a + 4d...
\boxed{\longmapsto \rm{a + a + d + a + 2d + a + 3d + a + 4d= 80}}⟼a+a+d+a+2d+a+3d+a+4d=80
\longmapsto \rm{5a + 10d = 80}⟼5a+10d=80
\longmapsto \rm{5(14)+ 10d = 80 \: \: \: [from \: (1)]}⟼5(14)+10d=80[from(1)]
\longmapsto \rm{70 + 10d = 80}⟼70+10d=80
\longmapsto \rm{10d = 80 - 70}⟼10d=80−70
\longmapsto \rm{10d = 10}⟼10d=10
\longmapsto \rm{d = \dfrac{ \cancel{10}}{ \cancel{10}} }⟼d=1010
\longmapsto \rm{ \green{d = 1}}⟼d=1
\rm{\therefore a {\tiny20}} = a + 19d∴a20=a+19d
\rm{\longmapsto a {\tiny20}} = 14 + 19(1)⟼a20=14+19(1)
\rm{\longmapsto a {\tiny20}} = 14 + 19⟼a20=14+19
\rm{\longmapsto \orange{ a {\tiny20}}}\orange{ = 33}⟼a20=33
Therefore, the 20th term of the A.P is 33.