Math, asked by KripanshPandey, 8 months ago

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.​

Answers

Answered by sweety2904
0

Step-by-step explanation:

Given third term is 16

T3 = 16

a+(n-1)d = 16

Here n = 3

a is the first term

d is the difference between two terms

a+(3-1)d = 16

a+2d = 16............ (1)

Given that the 7th term exceeds the 5th term by 12

So T 7 =T5 + 12

a+6d = a+4d+12

a+6d-a-4d-12=0

2d-12 = 0

d = 6

From eqn (1)..

a+2d= 16

a+2 (6)=16

a+ 12 = 16

a= 4

The AP are

4,4+6,4+2 (6),4+3 (6),4+4 (6).......

4,10,16,22,28,34,40..........

Answered by sourya1794
1

\bf{\underline{Solution}}:-

Let first term of AP be a

and common difference be d

Then,

\rm\:a_3=16

a + 2d = 16

a = 16 - 2d ..............................(i)

According to the question,

\rm\:a_7=a+4d+12

a + 6d = a + 4d + 12

6d = 4d + 12

6d - 4d = 12

2d = 12

d = \rm\cancel\dfrac{12}{2}

d = 6

putting the value of d in eq(i)

a = 16 - 2d

a = 16 - 2 × 6

a = 16 - 12

a = 4

Now,

\rm\:a_1=4

\rm\:a_2=a+d

\rm\longrightarrow\:a_2=4+6

\rm\longrightarrow\:a_2=10

\rm\:a_3=a+2d

\rm\longrightarrow\:a_3=4+2\times{6}

\rm\longrightarrow\:a_3=4+12

\rm\longrightarrow\:a_3=16

\rm\:a_4=a+3d

\rm\longrightarrow\:a_4=4+3\times{6}

\rm\longrightarrow\:a_4=4+18

\rm\longrightarrow\:a_4=22

Hence,AP will be 4,10,16,22,......

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