Question

Find the 98th term of the arithmetic sequence -10, -8, -6

Answer 1

Answer:

here the given sequence is -10,-8,-6....

a=-10. a2=-8 a3=-6.

d=a2-a1

d=-8-(-10)

d=-8+10

d=2.

d=a3-a2

d=-6-(-8)

d=-6+8

d=2

here, a1=-10, a2=-8, a3=-6, n=98, d=2

Now, an=a+(n-1)d

a98=-10+(98-1)2

a98=-10+97×2

a98= -10+194

a98= 184

So, the 98th term of the given A.P. will be 18

thank you

Answer 2

Answer:

Required 98th term is 184

Step-by-step explanation:

Given series is -10, -8, -6

First term of the series is 255 and common difference is (-8-(-10)) = 2

Here we want to find 98 th term of the series.

We know,

$a_n = a + (n - 1)d$

Where $a_n$ is nth term.

n is number of term and a is first term of the series and d is common difference.

Here,

$a = - 10 \\ d = 2 \\ n = 98$

So,

$a_{98} = - 10 + (98 - 1) \times ( 2) \\ = - 10 + 97 \times 2 \\ = - 10 + 194 \\ = 184$

Required 98 th term is 184

Here applied formula,$a_n = a + (n - 1)d$

One more important formula of AP series,

Sum of the series,

$s_n = \frac{n}{2} ( first \: term + last \: term)$

This is a problem of Arithmetic progression.

Know more about Arithmetic progression

1) https://brainly.in/question/4219484

2) https://brainly.in/question/2768711

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