Math, asked by shahharshitmodel, 1 year ago

If five times the fifth term of an A.P is equal to 8times its eigth term show that its 13th is zero

Answers

Answered by KunalVerma911
1
Hello Harshit
let the first term of the AP be a, and common difference be d.
Then nth term,An=a+(n-1)d
Now according to question,
5 x 5th term= 8 x 8th term
or,5 x (a+(5-1)d)=8x (a+(8-1)d)
or,5 x(a+4d)=8 x (a+7d)
or,5a+20d=8a+56d
or,5a-8a=36d
or,-3a=36d
or,a=-12d. (1)
Now 13 th term will be
a+(13-1)d
or,a+12d
or,a-a=0. (from 1)

shahharshitmodel: Thanks
KunalVerma911: Mention not
Answered by TRISHNADEVI
1

 \huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \:   \: QUESTION \:  \: } \mid}}}}}

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \text{If 5 times the 5th term of an AP is equal } \\ \text{to 8 times its 8th term, show that the 13th} \\  \text{ term is zero.}</p><p>

 \huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \:   \: SOLUTION \:  \: } \mid}}}}}

 \underline{ \mathfrak{  \:  \: Given, \:  \: }} \\\  \\  \mathtt{ \rightsquigarrow  5 \:  times \:  \:  of  \:  \: the  \:  \: 5th  \:  \: term = 8th \:  \:times} \\ \mathtt{ \: \: \: \:  \:\: \: \: \: \:\: \: \: \: \: \:\: \: \: \: \: \: \: \: \:\: \: \:\: \: \:\: \:  \: \: of  \:  \: the  \:  \: 8th  \:  \: term} \\  \\  \\  \underline{ \mathfrak{ \:  \: </p><p>To  \: show : \mapsto \:  \: }} \\  \\  \mathtt{ \rightsquigarrow \: 13th  \:  \: term \:  \:  is \:  \:  equal  \:  \: to \:  \:  zero. \: }

 \underline{ \mathfrak{ \:  \: Suppose, \:  \: }} \\   \\ \mathtt{ \rightsquigarrow \: First  \:  \: term  \:  \: of  \:  \: the  \:  \: A.P. = a} \\  \mathtt{ \rightsquigarrow \: Difference = d}

 \underline{ \bold{ \:  A.T.Q.,  \: }} \\  \\ \:  \:  \:  \:  \:  \:  \mathtt{ 5  \times T_5 = 8 \times  T_8 }\\  \\ \mathtt{\Longrightarrow \:  5[ a + (5-1)d]=8 [ a + (8-1)d]} \\  \\   \mathtt{\Longrightarrow 5(a  + 4d)= 8 (a+7d) }\\  \\    \mathtt{ \Longrightarrow 5a + 20 \: d = 8a + 56 \: d} \\  \\  \mathtt{  \Longrightarrow 5a - 8a = 56 \: d    - 20 \: d}\\  \\  \mathtt{\Longrightarrow - 3a   =  36 \: d}\\  \\   \mathtt{\Longrightarrow  a =  \frac{36 \: d}{ -  3}}  \\  \\  \:  \:  \:  \mathtt{ \therefore \:  \: \underline{  \: a = -  12 \: d \: }}</p><p></p><p>

  \underline{\mathfrak{ \: Now \: }} \\  \\ \mathtt {13th  \:  \: term,  \: T_{13 }= a + [(13-1)d] } \\  \\    \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \mathtt{ = a + 12 \: d} \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \mathtt{ = ( - 12 \: d) + 12 \: d } \\  \mathtt{\:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:   \red{[As \:  \: a  =  - 12 \: d]}} \\  \\    \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:\mathtt{ =  - 12 \: d + 12 \: d} \\  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \: \:  \mathtt{ = 0} \\  \\  \\    \huge{\mathtt{ \therefore \:  \:  \underline{ \:  \: 13th  \:  \: term \:  = 0 \:  \: }} }\\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: </p><p> \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \mathtt{ \underline{ \: Hence \:  \: proved. \:  \: }}

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