Question

Consider the sequence seventh term is 34 fifth term is 64. Find common difference find twentieth term

Answer 1

Answer:

Common difference=-15 and twentieth term is -41

Step-by-step explanation:

Given seventh term(a₇)=34

a+6d=34  ---->(1)

Fifth term(a₅)=64

a+4d=64--->(2)

eq(1)-eq(2)

⇒a+6d-a-4d=34-64

⇒2d=-30

⇒d=-15

Common difference=-15

substitute d=-15 in eq(2)

⇒a+4(-15)=64

⇒a-60=64

⇒a=124

First term a=124

Twentieth term a₁₂=a+11d

                              =124+11(-15)

                              =-41

Answer 2

Given :-

Seventh term of the AP = 34

Fifth term of the AP = 64

Required to find :-

• Common difference ?

• 20th term ?

Formulae used :-

$\huge{\dagger{\boxed{\rm{ {a}_{nth} = a + \big( n - 1 \big) d }}}}$

Solution :-

Given information :-

7th term of the AP = 34

5th term of the AP = 64

we need to find the common difference and 12th term .

So,

Let's consider the given information ;

7th term = 34

we know that

7th term can be represented as " a + 6d "

So,

a + 6d = 34 $\longrightarrow{\text{Equation-1}}$

consider this as equation - 1

Similarly,

5th term = 64

5th term can be represented as " a + 4d "

So,

a + 4d = 64 $\longrightarrow{\text{Equation-2}}$

consider this as equation - 2

Now,

we need to solve the 2 linear equations simultaneously .

So,

Subtract equation 2 from equation 1

$\tt a + 6d = 34 \: \\ \tt a + 4d = 64 \\ \tt (-)(-) (-) \\ \rule{65}1 \\ \cancel{0} + 2d = - 30 \\ \rule{65}1 \\ 2d = - 30 \\ \\ d = \displaystyle \dfrac{ - 30}{2} \\ \\ \tt d \: = - 15$

So,

Value of " d " = - 15

Common difference = - 15

However,

Substitute the value of d in equation 1

a + 6d = 34

a + 6 ( - 15 ) = 34

a + ( - 90 ) = 34

a - 90 = 34

a = 34 + 90

a = 124

So,

value of " a " is 124

Using the formula ,

$\huge{\dagger{\boxed{\rm{ {a}_{nth} = a + \big( n - 1 \big) d }}}}$

This formula enables us to find the 20th term

So,

here

$\boxed{\boxed{\begin{minipage}{6.4cm} \\ \bf{ a = first term } \\ \bf{d = common difference} \\ \bf{n = the term number which you want to find} \end{minipage}}}$

Hence,

$\rightarrow{\rm{ {a}_{nth} = {a}_{20} }}$

$\rightarrow{\rm{ {a}_{20} = 124 + ( 20 - 1 ) - 15 }}$

$\rightarrow{\rm{ {a}_{20} = 124 + ( 19 ) - 15 }}$

$\rightarrow{\rm{{a}_{20} = 124 + ( - 285 ) }}$

$\rightarrowtail{\rm{ {a}_{20} = - 161 }}$

20th term = - 161