Sir Rod Stewart will be joined at Glastonbury by Ronnie Wood.
The ‘Maggie May’ hitmaker will play he prestigious Sunday afternoon Legends slot at the famous festival in Somerset, south west England, in June and he’s planning to give a nod to the early days of his career by welcoming his friend and former Faces bandmate to the stage at Worthy Farm.
A source told The Sun on Sunday newspaper’s Bizarre column: “Rod is really excited about his big Worthy Farm gig and wanted to give a nod to the beginning of his career, so thought it would be fitting to bring out Ronnie.
“He knows the pair of them together will be a huge crowd pleaser.”
Rod previously revealed that, although he jumped at the chance to play Glastonbury, it will cost him a lot of money.
He then added: “It’s going to cost me a fortune to do it – $300,0000.
“I’ve got to bring all my band back from America, of course Glastonbury don’t pay for that. But I don’t care if it cost me $1 million, I would have done it. This is a great honour. It really is the greatest honour.”
The 80-year-old star previously declared that he is “proud, ready and more than able to pleasure and titillate my friends at Glastonbury” next summer.
He wrote on Instagram: “I’m absolutely thrilled to announce that I’ll be playing Glastonbury Festival 2025! After all these years, I’m proud and ready and more than able to take the stage again to pleasure and titillate my friends at Glastonbury in June. I’ll see you there!”
Glastonbury co-organiser Emily posted: “Bringing Sir Rod Stewart back for the Sunday afternoon slot on the Pyramid Stage is everything we could wish for.
“What a way to bow out with the final legends slot before we take a fallow year. We cannot wait.”
Glastonbury will take a year off in 2026 for the traditional fallow year reserved for the rehabilitation of the land at Worthy Farm.
Rod will become the first artist to have headlined Glastonbury and later been given legends billing after he previously topped the bill at the festival in 2002 alongside Coldplay and Stereophonics.