Question

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Answer 1

Let 1st term is ‘a’ & common difference is ‘d’

n^th term =a+( n-1)d

12th term =a +11d=(-13).......... eq^n.. 1

Now, sum of 1st four term,

(a)+(a+d)+(a+2d)+(a+3d)=24

» 4a+6d=24

» 2a+3d=12...... eq^n... 2

Now , eq^n.. 1×2

» 2a+22d=(-26)

Now you have two variable & two eq, so solve this & find the value of a & d then

Use the formula

Sum of n^th term = (n/2) { 2a+(n-1)d}

Put the value of n=10 & the value of ‘a’& ‘d’ that you get after solving both equation

Answer 2

Step-by-step explanation:

a12 = -13

a+ (12-1)d

a+11d = -13 -(1)

S4 = 4/2{2a+(4-1)d}

24 = 2(2a+3d)

24/2 = 2a+3d

2a+3d = 12

Multiplying eq 1 by 2, we get

2a+22d = -26

Subtracting eq 1 from 2, we get

2a+3d =12

2a+22d = -26

- - +

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-19d = 38

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d = -2