Math, asked by natasha23, 1 year ago

The fourth term of an AP is 11 and the eighth term exceeds twice the fourth term by 5. Find the AP and the sum of first 50 terms

Answers

Answered by neharajput12
105
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Answered by wifilethbridge
23

Answer:

Step-by-step explanation:

Sequence of AP : a,a+d,a+2d,.....

Formula of nth term in AP = a+(n-1)d

We are given that The fourth term of an AP is 11

a_4=11

So, put n = 4 in the formula

a_4=a+(4-1)d

a_4=a+3d

11=a+3d ---1

Since we are given that  the eighth term exceeds twice the fourth term by 5.

a_8=2(a_4)+5

Using formula of nth term .

a+(8-1)d=2(a_4)+5

a+7d=2*11+5

a+7d=27--2

Subtract 1 from 2

a+7d-(a+3d)=27-11

a+7d-a-3d=27-11

7d-3d)=16

4d=16

d=4

Putting value of d in equation 1 to get the value of a

11=a+3*4

a=-1

Now putting values of a and d in the general form of AP :

-1, -1+4, -1+2*4,.....

-1, 3,7.....

Thus the sequence is -1, 3,7.....

Formula of sum of n terms in AP : \frac{n(2a+(n-1)d}{2}

Since n = 50

Sum of first 50 terms in this AP = \frac{50(2*-1+(50-1)*4}{2}

                                                        =\frac{50(-2+49*4)}{2}

                                                        =\frac{50(-2+196)}{2}

                                                        =\frac{50(194)}{2}

                                                        =\frac{9700}{2}

                                                         =4850

Hence the sum of first 50 terms is 4850

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