The sum of three consecutive terms of an A.P. is 36 and their product is 1140.
Find the terms. (Consider the terms to be in d

Question

The sum of three consecutive terms of an A.P. is 36 and their product is 1140. Find the terms. (Consider the terms to be in descending order.)​

Eliza · Answer

Answer:     Let         the         small    est         Num    ber     of         A    .    P         be         x    .           so    ,         the         oth    er         numb    ers      =    >         x    +    1         a    nd         x+    2           Giv    en     th    at    ,           sum         of         these         te    rms         =         3    6           Acc    ording     to     the         questions      x+    2    +    x+    1    +    x         =         3    6           =    >         3    x    +    3         =         3    6           =    >         3    x         =         3    6    -    3           =    >         3    x         =         3    3           =    >         x         =         3    3    /    3           =    >         x         =         1    1           Ther    efore         the         term    s         are    ,           =    >         x    +    2         =         1    1    +    2         =         1    3           =    >         x    +    1         =         1    1    +    1         =         1    2           =    >         x         =         1    1

Emery · Answer

Answer:

5,12,19

Step-by-step explanation:

let a-d , a , a+d be the 3 terms of A.P such that

a-d + a + a+d = 36 (given)

=> 3a = 36

=> a = 36/3

=> a= 12

also given (a-d)* a * (a+d) = 1140

=> (12-d) * (12) * (12+d) = 1140

=> (12-d) (12+d) = 1140/12

=> 12²- d² = 95

=> d² = 144 – 95 = 49

=> d = √49 = 7

=> the 3 terms are , a-d, a, a+d

12-7, 12,12+7

5,12,19

Hope this help you

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Rose

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The sum of three consecutive terms of an A.P. is 36 and their product is 1140.Find the terms. (Consider the terms to be in d