the sum of three consecutive terms in ap is 6 and their products is -120 find the three numbers ( hint let 3 numbers be a-d ,a,a+d
Answers
In AS( arithmetic sequence ),
ғɪʀsᴛ ᴛᴇʀᴍ = ᴀ
ᴄᴏᴍᴍᴏɴ ᴅɪғғᴇʀᴇɴᴄᴇ = ᴅ
sᴇᴄᴏɴᴅ ᴛᴇʀᴍ = ᴀ₂ = ᴀ + ᴅ
ᴛʜɪʀᴅ ᴛᴇʀᴍ = ᴀ₃ = ᴀ + 2ᴅ
xᴛʜ ᴛᴇʀᴍ = ᴀₓ = ᴀ + ( x - 1 )ᴅ
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ʟᴇᴛ ᴛʜᴇ ᴄᴏɴsᴇᴄᴜᴛɪᴠᴇ ᴛᴇʀᴍs ᴏғ ᴀs ʙᴇ ( ᴀ - ᴅ ) , ᴀ , ( ᴀ + ᴅ ).
In the question it is given that the sum of the consecutive terms of that AS is 12.
⇒ ( ᴀ - ᴅ ) + ᴀ + ( ᴀ + ᴅ ) = 12
⇒ ᴀ - ᴅ + ᴀ + ᴀ + ᴅ = 12
⇒ ᴀ + ᴀ + ᴀ - ᴅ + ᴅ = 12
⇒ 3ᴀ = 12
⇒ ᴀ = 12 / 3
⇒ ᴀ = 4
Also given that the product of the same terms in the AS is 48.
⇒ ( ᴀ - ᴅ ) ( ᴀ ) ( ᴀ + ᴅ ) = 48
⇒ ( ᴀ - ᴅ ) ( ᴀ + ᴅ ) ᴀ = 48
⇒ ( ᴀ^2 - ᴅ^2 ) ᴀ = 48
sᴜʙsᴛɪᴛᴜᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ ᴀ :
⇒ ( 4^2 - ᴅ^2 ) 4 = 48
⇒ 16 - ᴅ^2 = 48 / 4
⇒ 16 - ᴅ^2 = 12
⇒ 16 - 12 = ᴅ^2
⇒ 4 = ᴅ^2
⇒ \ᴘᴍ 2 = ᴅ±2=ᴅ
⇒ 2 ᴏʀ - 2 = ᴅ
ɴᴏᴡ, ᴛʜᴇʀᴇ ᴀʀᴇ ᴛᴡᴏ ᴠᴀʟᴜᴇs ᴏғ ᴅ ᴏʀ ᴄᴏᴍᴍᴏɴ ᴅɪғғᴇʀᴇɴᴄᴇ, sᴜʙsᴛɪᴛᴜᴛɪɴɢ ʙᴏᴛʜ ᴛʜᴇ ᴠᴀʟᴜᴇs ᴏғ ᴅ ɪɴ ᴛᴇʀᴍs ᴏғ ᴀᴘ ᴏɴᴇ ʙʏ ᴏɴᴇ.
If we take the value of d equal to 2. Arithmetic progressions are :
ᴀ - ᴅ = 4 - 2 = 2
ᴀ = 4
ᴀ + ᴅ = 4 + 2 = 6
If we take the value of d equal to - 2. Arithmetic progressions are :
ᴀ - ᴅ = 4 - ( - 2 ) = 4 + 2 = 6
ᴀ = 4
ᴀ + ᴅ = 4 + ( - 2 ) = 4 - 2 = 2
Answer:
3
Step-by-step explanation:
(a-d) +a +(a+d) =6
now remove the brackets
a-d+a+a+d=6
-d+d are striked
a+a+a =6
3a=6
a=6/3
a=2
(a-d)*a*(a+d) =120
let a be 2
(2-d)*2*(2+d)=120
(2-d)(2+d)*2=120
4 +2d -2d +dd =120
4+dd=120
dd=120-4
√dd=116