Question

t3+t7=6 and t3.t7=8 then find a and d

Answer 1

Answer:

Step-by-step explanation:

t3= a+2d

t7= a+6d

t3+ t7 = 6

⇒a+2d+a+6d = 6

⇒2a+ 8d =6

⇒a = 6-8d / 2

⇒a= 3- 4d ---------------(i)

Now,

(a+2d)(a+ 6d) = 8

⇒a² +6da +2da +12d²=8

apply (i),

(3-4d)² + 8d (3-4d)+ 12 d²=8

⇒9+16d²_24d+24d-32d²+12d²=8

⇒-4d² +9 =8

⇒-4d²= -1

⇒d²= 1/4

⇒d=1/2 ---------------(ii)

Apply (ii) in (i):-

a= 3-4d

⇒a=3- 4× 1/2

⇒a =1

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Answer 2

Answer:

If t3+t7=6 and t3.t7=8 then  a and d are 1, 1/2 respectively.

Step-by-step explanation:

By considering the arithmetic progression, t3 represents third term and t7 represents the seventh term of an arithmetic progression.

What is Arithmetic Progression?

The sequence of numbers in which the consecutive numbers have a difference that is constant called the common difference,d.

Terms in Arithmetic Progression are as a, a+d, a+2d, a+3d.......

$n^{th}$ term of an arithmetic progression is a + (n-1)d.

Calculation:

Third term of Arithmetic Progression, $t_{3}$ = a+2d

Seventh term of Arithmetic Progression, $t_{7}$ = a+6d

$t_{3}+t_{7} = 6\\a+2d+a+6d = 6\\2a+8d = 6\\a+4d = 3$ .....(1)

$t_{3} *t_{7} = 8\\(a+2d)*(a+6d) = 8$......(2)

From (1), a = 3 - 4d   ......(3)

Substitute the value of a in (2)

$(3-4d+2d)*(3-4d+6d) = 8\\(3-2d)*(3+2d) = 8\\9-4d^{2} =8\\4d^{2} = 1\\d = \frac{1}{2}$

Substituting the value of d in (3)

$a = 3-4(\frac{1}{2} )\\a = 3-2\\a = 1$

Hence, the values of a and d are 1, 1/2 respectively.

Find more about Progressions:

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