1 thought on “ques. Tʜᴇ ᴀɴɢʟᴇ ᴏғ ᴇʟᴇᴠᴀᴛɪᴏɴ ᴏғ ᴛʜᴇ ᴛᴏᴘ ᴏғ ᴀ ᴛᴏᴡᴇʀ ғʀᴏᴍ ᴛᴡᴏ ᴘᴏɪɴᴛs ᴅɪsᴛᴀɴᴛ s ᴀɴᴅ ᴛ ғʀᴏᴍ ɪᴛs ғᴏᴏᴛ ᴀʀᴇ ᴄᴏᴍᴘʟᴇᴍᴇɴᴛᴀʀʏ. Pʀᴏᴠᴇ ᴛʜᴀᴛ ᴛʜᴇ”
Answer:
Given, first term, a = 10
361 = 10 + (n − 1)9361 = 10 + 9n − 9
Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 1
Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360
Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40
Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP = 40
Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP = 40Now, sum of total number of terms of an AP is given as:
Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP = 40Now, sum of total number of terms of an AP is given as:Sn = n/2 [2a + (n − 1)d]
Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP = 40Now, sum of total number of terms of an AP is given as:Sn = n/2 [2a + (n − 1)d]S40 = 40/2 [2 x 10 + (40 − 1)9]
Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP =
Answer:
Given, first term, a = 10
361 = 10 + (n − 1)9361 = 10 + 9n − 9
Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 1
Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360
Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40
Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP = 40
Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP = 40Now, sum of total number of terms of an AP is given as:
Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP = 40Now, sum of total number of terms of an AP is given as:Sn = n/2 [2a + (n − 1)d]
Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP = 40Now, sum of total number of terms of an AP is given as:Sn = n/2 [2a + (n − 1)d]S40 = 40/2 [2 x 10 + (40 − 1)9]
Given, first term, a = 10Last term, al = 361And, common difference, d = 9al = a + (n −1)d361 = 10 + (n − 1)9361 = 10 + 9n − 9361 = 9n + 19n = 360n = 40Therefore, total number of terms in AP =