Question

in an arithmetic progression (A.P.) the fouth term and sixth terms are 18 and 14 respectively.
Find the i) first team ii) common difference iii) sum of first 20 terms ​

Answer 1

Step-by-step explanation:

t4 = 18

a+(n-1)d = 18

a +3d = 18............(1)

t6 = 14

a+(n-1)d = 14

a +5d = 14........(2)

Let's solve simultaneously (1)&(2)

a + 3d = 18

a + 5d = 14

_____________

0 -2d = 4

(d = -2) = common difference

so let's fine a = t1 = first term

a + 3d = 18

a= 18 +6

a = (24) = first term

Sum of first 20 terms

S20 = n/2 [2a +(n-1) d]

S20 = 20/2 [ 48 + 19×(-2) ]

S20 = 20/2 [ 48 + -38]

S20 = 20/2 [10 ]

S20 = 100

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Hope it helps you buddy

Answer 2

Answer:

a = 24

d = - 2

S₂₀ = 100

Step-by-step explanation:

Given ;

$\displaystyle{t_4=18}\\\\\\\displaystyle{t_6=14}\\\\\\\displaystyle{We \ know \ term \ formula}\\\\\\\displaystyle{t_n=a+(n-1)d}\\\\\\\displaystyle{t_4=a+(4-1)d}\\\\\\\displaystyle{18=a+3d}\\\\\\\displaystyle{a=18-3d \ ..(i)}\\\\\\\displaystyle{t_6=a+(6-1)d}\\\\\\\displaystyle{14=a+5d}\\\\\\\displaystyle{a=14-5d \ ..(ii)}$

From ( i )  and  ( ii )  we have

18 - 3 d = 14 - 5 d

- 3 d + 5 d = 14 - 18

d = - 2

So ,  common difference is - 2

putting d = - 2 in ( i )

a = 18 - 3 d

a = 24

So ,  first term is 24 .

We know sum formula

$\displaystyle{S_n=\dfrac{n}{2}\left[2a+(n-1)d\right]}\\\\\\\displaystyle{S_{20}=\dfrac{20}{2}\left[2\times24+(20-1)-2\right]}\\\\\\\displaystyle{S_{20}=10[48-38]}\\\\\\\displaystyle{S_{20}=10[10]}\\\\\\\displaystyle{S_{20}=100}$

Sum of first 20 terms ​is 100 .