find the sum of first 40 positive integers divisible by 6.
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ᴛʜᴇ ғɪʀsᴛ 40 ᴘᴏsɪᴛɪᴠᴇ ɪɴᴛᴇɢᴇʀs ᴅɪᴠɪsɪʙʟᴇ ʙʏ 6 ᴀʀᴇ 6,12,18,....... ᴜᴘᴛᴏ
40 ᴛᴇʀᴍs
ᴛʜᴇ ɢɪᴠᴇɴ sᴇʀɪᴇs ɪs ɪɴ ᴀʀᴛʜɪᴍᴇᴛɪᴄ ᴘʀᴏɢʀᴇssɪᴏɴ ᴡɪᴛʜ ғɪʀsᴛ ᴛᴇʀᴍ ᴀ=6 ᴀɴᴅ ᴄᴏᴍᴍᴏɴ ᴅɪғғᴇʀᴇɴᴄᴇ ᴅ=6
sᴜᴍ ᴏғ ɴ ᴛᴇʀᴍs ᴏғ ᴀɴ ᴀ.ᴘ ɪs
ɴ/2×{2ᴀ+(ɴ-1)ᴅ}
→ʀᴇϙᴜɪʀᴇᴅ sᴜᴍ = 40/2×{2(6)+(39)6}
=20{12+234}
=20×246
=4920
ɪ ʜᴏᴘᴇ ᴛʜɪs ᴡɪʟʟ ʜᴇʟᴘ ᴜ ;)
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2
AP for the above is = 6,12,18,........
a= 6
d= 6
40th term= a+39d
= 6+ 39*6
= 240
sum of the first 40 positive integers divisible by 6 is = n/2{2a+(n-1)d}
= 20(12+234)
= 20 * 246
= 4920
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